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Question Number 117708 by mr W last updated on 13/Oct/20

Find the number of all 5 digit numbers x_1 x_2 x_3 x_4 x_5 with x_1 ≥x_2 ≥x_3 ≥x_4 ≥x_5 .

$${Find}\:{the}\:{number}\:{of}\:{all}\:\mathrm{5}\:{digit} \\ $$$${numbers}\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} {x}_{\mathrm{4}} {x}_{\mathrm{5}} \:{with} \\ $$$${x}_{\mathrm{1}} \geqslant{x}_{\mathrm{2}} \geqslant{x}_{\mathrm{3}} \geqslant{x}_{\mathrm{4}} \geqslant{x}_{\mathrm{5}} . \\ $$

Commented by prakash jain last updated on 13/Oct/20

(Σ_(j=0) ^4 ^4 C_j ×^(10) C_(j+1) )−1 ^4 C_j indicates partion of 5 into j+1 parts. ^(10) C_(j+1) way to choose j+1 digits. −1 to exclude all zero cases for n digits (Σ_(j=0) ^(n−1) ^(n−1) C_j ×^(10) C_(j+1) )−1

$$\left(\underset{{j}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\:^{\mathrm{4}} {C}_{{j}} ×^{\mathrm{10}} {C}_{{j}+\mathrm{1}} \right)−\mathrm{1} \\ $$$$\:^{\mathrm{4}} {C}_{{j}} \:\mathrm{indicates}\:\mathrm{partion}\:\mathrm{of}\:\mathrm{5}\:\mathrm{into}\:{j}+\mathrm{1}\:\mathrm{parts}. \\ $$$$\:^{\mathrm{10}} {C}_{{j}+\mathrm{1}} \:\mathrm{way}\:\mathrm{to}\:\mathrm{choose}\:{j}+\mathrm{1}\:\mathrm{digits}. \\ $$$$\:\:−\mathrm{1}\:\mathrm{to}\:\mathrm{exclude}\:\mathrm{all}\:\mathrm{zero}\:\mathrm{cases} \\ $$$$\mathrm{for}\:{n}\:\mathrm{digits} \\ $$$$\left(\underset{{j}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\:\:^{{n}−\mathrm{1}} {C}_{{j}} ×^{\mathrm{10}} {C}_{{j}+\mathrm{1}} \right)−\mathrm{1} \\ $$

Commented by prakash jain last updated on 13/Oct/20

Distribution calculation 5 digits • • • • • We can 5 digits in j+1 number of part by placing j bar in space between dots. for a 5 digit number formed with 2 unique digits we have following ways 1d_1 +4d_2 =•∣•••• 1d_1 +4d_2 ,2d_1 +3d_2 ,3d_1 +2d_2 ,4d_1 +d_2 = ^4 C_1 there are^(10) C_2 ways to select 2 digits. similar for j digits ^4 C_j ways to decide which digit is repeated how many time. ^(10) C_j ways to select j digits. once we have selected digits and decided how many times each digit is repeated then there is only one unique way of writing sorted number. so numbers of way a 5 digit number can be written using j different digits in the required sorted order is. ^4 C_(j−1) ×^(10) C_j we need to subtract 1 for case where a single digit is repeated 5 times. so 5 digit numbers in required sorted order are 2001.

$$\mathrm{Distribution}\:\mathrm{calculation} \\ $$$$\mathrm{5}\:\mathrm{digits} \\ $$$$\bullet\:\:\bullet\:\:\bullet\:\:\bullet\:\:\bullet \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{5}\:\mathrm{digits}\:\mathrm{in}\:{j}+\mathrm{1}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{part}\:\mathrm{by}\:\mathrm{placing}\:{j}\:\mathrm{bar}\:\mathrm{in}\:\mathrm{space}\:\mathrm{between} \\ $$$$\mathrm{dots}. \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{5}\:\mathrm{digit}\:\mathrm{number}\:\mathrm{formed}\:\mathrm{with} \\ $$$$\mathrm{2}\:\mathrm{unique}\:\mathrm{digits}\:\mathrm{we}\:\mathrm{have}\:\mathrm{following} \\ $$$$\mathrm{ways} \\ $$$$\mathrm{1}{d}_{\mathrm{1}} +\mathrm{4}{d}_{\mathrm{2}} =\bullet\mid\bullet\bullet\bullet\bullet \\ $$$$\mathrm{1d}_{\mathrm{1}} +\mathrm{4d}_{\mathrm{2}} ,\mathrm{2}{d}_{\mathrm{1}} +\mathrm{3}{d}_{\mathrm{2}} ,\mathrm{3}{d}_{\mathrm{1}} +\mathrm{2}{d}_{\mathrm{2}} ,\mathrm{4}{d}_{\mathrm{1}} +{d}_{\mathrm{2}} \\ $$$$=\:\:^{\mathrm{4}} {C}_{\mathrm{1}} \\ $$$$\mathrm{there}\:\mathrm{are}\:^{\mathrm{10}} {C}_{\mathrm{2}} \:\mathrm{ways}\:\mathrm{to}\:\mathrm{select}\:\mathrm{2}\:\mathrm{digits}. \\ $$$$ \\ $$$$\mathrm{similar}\:\mathrm{for}\:{j}\:\mathrm{digits} \\ $$$$\:^{\mathrm{4}} {C}_{{j}} \:\mathrm{ways}\:\mathrm{to}\:\mathrm{decide}\:\mathrm{which}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{repeated} \\ $$$$\:\:\mathrm{how}\:\mathrm{many}\:\mathrm{time}. \\ $$$$\:\:^{\mathrm{10}} {C}_{{j}} \:\mathrm{ways}\:\mathrm{to}\:\mathrm{select}\:{j}\:\mathrm{digits}. \\ $$$$\mathrm{once}\:\mathrm{we}\:\mathrm{have}\:\mathrm{selected}\:\:\mathrm{digits} \\ $$$$\mathrm{and}\:\mathrm{decided}\:\mathrm{how}\:\mathrm{many}\:\mathrm{times} \\ $$$$\mathrm{each}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{repeated}\:\mathrm{then} \\ $$$$\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:\mathrm{unique}\:\mathrm{way}\:\mathrm{of} \\ $$$$\mathrm{writing}\:\mathrm{sorted}\:\mathrm{number}. \\ $$$$\mathrm{so}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{way}\:\mathrm{a}\:\mathrm{5}\:\mathrm{digit}\:\mathrm{number} \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{using}\:{j}\:\mathrm{different}\:\mathrm{digits} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{required}\:\mathrm{sorted}\:\mathrm{order}\:\mathrm{is}. \\ $$$$\:^{\mathrm{4}} {C}_{{j}−\mathrm{1}} ×^{\mathrm{10}} {C}_{{j}} \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{subtract}\:\mathrm{1}\:\mathrm{for}\:\mathrm{case}\:\mathrm{where} \\ $$$$\mathrm{a}\:\mathrm{single}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{repeated}\:\mathrm{5}\:\mathrm{times}. \\ $$$$\mathrm{so}\:\mathrm{5}\:\mathrm{digit}\:\mathrm{numbers}\:\mathrm{in}\:\mathrm{required} \\ $$$$\mathrm{sorted}\:\mathrm{order}\:\mathrm{are}\:\mathrm{2001}. \\ $$

Commented by mr W last updated on 13/Oct/20

nice solution sir!

$${nice}\:{solution}\:{sir}! \\ $$

Answered by 1549442205PVT last updated on 14/Oct/20

For convenient we write abcde^(−) instead of x_1 x_2 x_3 x_4 x_5 ^(−) and ∣abcde∣−number of numbers of form abcde^(−) We solve the problem in turn for cases 1)How many are there the two−digit numbers ab^(−) can be formed from the digits 1,..,9 such that a≥b This case,it is easy to see in the total we can establish p=C_9 ^2 +9=45 ones 2)The case abc^(−) . When 9 at first place we have C_8 ^2 +8+9=45 numbers of form 9bc^(−) Similarly,∣8bc^(−) ∣= C_7 ^2 +7+8=36 ones ,.....,∣5bc^(−) ∣=C_4 ^2 +4+5=15 ones ∣4bc^(−) ∣=C_3 ^2 +3+4=10;∣3bc^(−) ∣=C_2 ^2 +2+3=6 ∣2bc∣=3,∣1bc∣=1 .Hence,in total we obtain ∣abc∣=∣9bc∣+∣8bc∣+∣7bc∣+6bc∣+∣5bc∣ +∣4bc∣+∣3bc∣+∣2bc∣+∣1bc∣= 45+36+28+21+15+10+6+4=165 numbers 3)The case abcd^(−) ∣abcd∣=∣9bcd∣+∣8bcd∣+...+∣3bcd∣+∣2bcd∣+∣1bcd∣ From above result we get: ∣9bcd∣=∣9cd∣+∣8cd∣...+∣1cd∣=45+36 28+21+15+10+6+4=165 ∣8bcd∣=∣8cd∣+...+∣1cd∣=165−45=120 ∣7bcd∣=∣8bcd∣−∣8cd∣=120−36=84 ∣6bcd∣=∣7bcd∣−∣7cd∣=84−28=56 ∣5bcd∣=∣6bcd∣−∣6cd∣=56−21=35 ∣4bcd∣=∣5bcd∣−∣5cd∣=35−15=20 ∣3bcd^(−) ∣= ∣4bcd∣−∣4cd∣=20−10=10 ∣2bcd∣=∣3bcd∣−∣3cd∣=10−6=4 ∣1bcd∣=∣2bcd∣−∣2cd∣=4−3=1 Hence,in total we obtain 165+120+84+56+35+20+10+5=495 4)The case abcde^(−) a)When 9 at first we have 9bcde^(−) From thepart 2)above we get ∣99cde^(−) ∣=∣9de∣+∣8de∣+∣7de∣+∣6de∣+∣5de∣+∣4de∣+∣3de∣ +∣5de∣+∣4de∣+∣3de∣+∣2de∣+∣1de∣ =45+36+28+21+15+10+6+4=165 ∣98cde∣=∣99cde∣−∣9de∣=165−45=120 ∣97cde∣=∣98cde∣−∣8de∣=120−36=84 ∣96cde∣=∣97cde∣−∣7de∣=84−28=56 ∣95cde∣=56−21=35,∣94cde∣=35−15=20 ∣93cde∣=10;∣92cde∣=4;∣91cde∣=1 .Hence,in the total we get 165+120+84+56+35+20+10+4+1 =495 numbers of forms 9bcde^(−) b)When 8 at the first place we have ∣8bcde^(−) ∣=∣88cde∣+∣87cde∣+...+∣81cde∣ =∣98cde∣+∣97cde∣+...+∣91cde∣ =∣9bcde∣−∣9cde∣=495−165=330 Brcause:See the formulas of part2) ∣88cde^(−) ∣=∣8de∣+∣7de∣+∣6de∣+∣5de∣ +∣4de∣+∣3de∣+∣2de∣+∣1de∣=165−45=120 ∣87cde∣=120−36=84 ∣86cde∣=84−28=56 ∣85cde∣=56−21=35 ∣84cde∣=35−15=20,∣83cde∣=10 ∣82cde∣=4,∣81cde∣=1,so in the total have 120+84+56+35+20+10+5=330 c)The case 7bcde^(−) we get (see the part3) ∣7bcde∣=∣8bcde∣−∣8cde∣ ∣=330−120=210 numbers d)The case 6bcde^(−) ;∣6bcde∣= ∣7bcde∣−∣7cde∣=210−84=126 e)The case5bcde^(−) ;∣ 5bcde^(−) ∣= ∣6bcde∣−∣6cde∣=126−56=70 f)The case 4bcde^(−) we get ∣4bcde∣=70−35=35 numbers g)The case 3bcde we get ∣3bcde∣=35−20=15 numbers ∣∣2bcde∣=15−10=5,∣1bcde∣=1 Finaly,in total we get 495+330+210+126+70+35+15+6= 1287 numbers of form abcde satisfy the condition a≥b≥c≥d≥e

$$\mathrm{For}\:\mathrm{convenient}\:\mathrm{we}\:\mathrm{write}\:\overline {\mathrm{abcde}}\:\mathrm{instead} \\ $$$$\mathrm{of}\overline {\:\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} \mathrm{x}_{\mathrm{3}} \mathrm{x}_{\mathrm{4}} \mathrm{x}_{\mathrm{5}} }\mathrm{and}\:\mid\mathrm{abcde}\mid−\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{numbers}\:\mathrm{of}\:\mathrm{form}\:\overline {\mathrm{abcde}} \\ $$$$\mathrm{We}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{in}\:\mathrm{turn}\:\mathrm{for}\:\mathrm{cases} \\ $$$$\left.\mathrm{1}\right)\mathrm{How}\:\mathrm{many}\:\mathrm{are}\:\mathrm{there}\:\mathrm{the}\:\mathrm{two}−\mathrm{digit}\:\:\mathrm{numbers}\: \\ $$$$\overline {\mathrm{ab}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{from}\:\mathrm{the}\:\mathrm{digits} \\ $$$$\mathrm{1},..,\mathrm{9}\:\mathrm{such}\:\mathrm{that}\:\mathrm{a}\geqslant\mathrm{b} \\ $$$$\mathrm{This}\:\mathrm{case},\mathrm{it}\:\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{in}\:\mathrm{the}\:\mathrm{total} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{establish}\:\mathrm{p}=\mathrm{C}_{\mathrm{9}} ^{\mathrm{2}} +\mathrm{9}=\mathrm{45}\:\mathrm{ones} \\ $$$$\left.\mathrm{2}\right)\mathrm{The}\:\mathrm{case}\overline {\:\mathrm{abc}}. \\ $$$$\mathrm{When}\:\mathrm{9}\:\mathrm{at}\:\mathrm{first}\:\mathrm{place}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{C}_{\mathrm{8}} ^{\mathrm{2}} +\mathrm{8}+\mathrm{9}=\mathrm{45}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{form}\:\overline {\mathrm{9bc}} \\ $$$$\mathrm{Similarly},\mid\overline {\mathrm{8bc}}\mid=\:\mathrm{C}_{\mathrm{7}} ^{\mathrm{2}} +\mathrm{7}+\mathrm{8}=\mathrm{36}\:\mathrm{ones} \\ $$$$,.....,\mid\overline {\mathrm{5bc}}\mid=\mathrm{C}_{\mathrm{4}} ^{\mathrm{2}} +\mathrm{4}+\mathrm{5}=\mathrm{15}\:\mathrm{ones} \\ $$$$\mid\overline {\mathrm{4bc}}\mid=\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{3}+\mathrm{4}=\mathrm{10};\mid\overline {\mathrm{3bc}}\mid=\mathrm{C}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}+\mathrm{3}=\mathrm{6} \\ $$$$\mid\mathrm{2bc}\mid=\mathrm{3},\mid\mathrm{1bc}\mid=\mathrm{1} \\ $$$$.\mathrm{Hence},\mathrm{in}\:\mathrm{total}\:\mathrm{we}\:\mathrm{obtain} \\ $$$$\mid\mathrm{abc}\mid=\mid\mathrm{9bc}\mid+\mid\mathrm{8bc}\mid+\mid\mathrm{7bc}\mid+\mathrm{6bc}\mid+\mid\mathrm{5bc}\mid \\ $$$$+\mid\mathrm{4bc}\mid+\mid\mathrm{3bc}\mid+\mid\mathrm{2bc}\mid+\mid\mathrm{1bc}\mid= \\ $$$$\mathrm{45}+\mathrm{36}+\mathrm{28}+\mathrm{21}+\mathrm{15}+\mathrm{10}+\mathrm{6}+\mathrm{4}=\mathrm{165}\:\mathrm{numbers}\: \\ $$$$\left.\mathrm{3}\right)\mathrm{The}\:\mathrm{case}\:\overline {\mathrm{abcd}} \\ $$$$\mid\mathrm{abcd}\mid=\mid\mathrm{9bcd}\mid+\mid\mathrm{8bcd}\mid+...+\mid\mathrm{3bcd}\mid+\mid\mathrm{2bcd}\mid+\mid\mathrm{1bcd}\mid \\ $$$$\mathrm{From}\:\mathrm{above}\:\mathrm{result}\:\mathrm{we}\:\mathrm{get}: \\ $$$$\mid\mathrm{9bcd}\mid=\mid\mathrm{9cd}\mid+\mid\mathrm{8cd}\mid...+\mid\mathrm{1cd}\mid=\mathrm{45}+\mathrm{36} \\ $$$$\mathrm{28}+\mathrm{21}+\mathrm{15}+\mathrm{10}+\mathrm{6}+\mathrm{4}=\mathrm{165} \\ $$$$\mid\mathrm{8bcd}\mid=\mid\mathrm{8cd}\mid+...+\mid\mathrm{1cd}\mid=\mathrm{165}−\mathrm{45}=\mathrm{120} \\ $$$$\mid\mathrm{7bcd}\mid=\mid\mathrm{8bcd}\mid−\mid\mathrm{8cd}\mid=\mathrm{120}−\mathrm{36}=\mathrm{84} \\ $$$$\mid\mathrm{6bcd}\mid=\mid\mathrm{7bcd}\mid−\mid\mathrm{7cd}\mid=\mathrm{84}−\mathrm{28}=\mathrm{56} \\ $$$$\mid\mathrm{5bcd}\mid=\mid\mathrm{6bcd}\mid−\mid\mathrm{6cd}\mid=\mathrm{56}−\mathrm{21}=\mathrm{35} \\ $$$$\mid\mathrm{4bcd}\mid=\mid\mathrm{5bcd}\mid−\mid\mathrm{5cd}\mid=\mathrm{35}−\mathrm{15}=\mathrm{20} \\ $$$$\:\mid\overline {\mathrm{3bcd}}\mid=\:\mid\mathrm{4bcd}\mid−\mid\mathrm{4cd}\mid=\mathrm{20}−\mathrm{10}=\mathrm{10} \\ $$$$\mid\mathrm{2bcd}\mid=\mid\mathrm{3bcd}\mid−\mid\mathrm{3cd}\mid=\mathrm{10}−\mathrm{6}=\mathrm{4} \\ $$$$\mid\mathrm{1bcd}\mid=\mid\mathrm{2bcd}\mid−\mid\mathrm{2cd}\mid=\mathrm{4}−\mathrm{3}=\mathrm{1} \\ $$$$\mathrm{Hence},\mathrm{in}\:\mathrm{total}\:\mathrm{we}\:\mathrm{obtain}\: \\ $$$$\mathrm{165}+\mathrm{120}+\mathrm{84}+\mathrm{56}+\mathrm{35}+\mathrm{20}+\mathrm{10}+\mathrm{5}=\mathrm{495} \\ $$$$\left.\mathrm{4}\right)\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{case}}\overline {\:\boldsymbol{\mathrm{abcde}}}\: \\ $$$$\left.\mathrm{a}\right)\mathrm{When}\:\mathrm{9}\:\mathrm{at}\:\mathrm{first}\:\mathrm{we}\:\mathrm{have}\:\overline {\mathrm{9bcde}} \\ $$$$\left.\mathrm{From}\:\mathrm{thepart}\:\mathrm{2}\right)\mathrm{above}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mid\overline {\mathrm{99cde}}\mid=\mid\mathrm{9de}\mid+\mid\mathrm{8de}\mid+\mid\mathrm{7de}\mid+\mid\mathrm{6de}\mid+\mid\mathrm{5de}\mid+\mid\mathrm{4de}\mid+\mid\mathrm{3de}\mid \\ $$$$+\mid\mathrm{5de}\mid+\mid\mathrm{4de}\mid+\mid\mathrm{3de}\mid+\mid\mathrm{2de}\mid+\mid\mathrm{1de}\mid \\ $$$$=\mathrm{45}+\mathrm{36}+\mathrm{28}+\mathrm{21}+\mathrm{15}+\mathrm{10}+\mathrm{6}+\mathrm{4}=\mathrm{165} \\ $$$$\mid\mathrm{98cde}\mid=\mid\mathrm{99cde}\mid−\mid\mathrm{9de}\mid=\mathrm{165}−\mathrm{45}=\mathrm{120} \\ $$$$\mid\mathrm{97cde}\mid=\mid\mathrm{98cde}\mid−\mid\mathrm{8de}\mid=\mathrm{120}−\mathrm{36}=\mathrm{84} \\ $$$$\mid\mathrm{96cde}\mid=\mid\mathrm{97cde}\mid−\mid\mathrm{7de}\mid=\mathrm{84}−\mathrm{28}=\mathrm{56} \\ $$$$\mid\mathrm{95cde}\mid=\mathrm{56}−\mathrm{21}=\mathrm{35},\mid\mathrm{94cde}\mid=\mathrm{35}−\mathrm{15}=\mathrm{20} \\ $$$$\mid\mathrm{93cde}\mid=\mathrm{10};\mid\mathrm{92cde}\mid=\mathrm{4};\mid\mathrm{91cde}\mid=\mathrm{1} \\ $$$$.\mathrm{Hence},\mathrm{in}\:\mathrm{the}\:\mathrm{total}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{165}+\mathrm{120}+\mathrm{84}+\mathrm{56}+\mathrm{35}+\mathrm{20}+\mathrm{10}+\mathrm{4}+\mathrm{1} \\ $$$$=\mathrm{495}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{forms}\:\overline {\mathrm{9bcde}} \\ $$$$\left.\mathrm{b}\right)\mathrm{When}\:\mathrm{8}\:\mathrm{at}\:\mathrm{the}\:\mathrm{first}\:\mathrm{place}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mid\overline {\mathrm{8bcde}}\mid=\mid\mathrm{88cde}\mid+\mid\mathrm{87cde}\mid+...+\mid\mathrm{81cde}\mid \\ $$$$=\mid\mathrm{98cde}\mid+\mid\mathrm{97cde}\mid+...+\mid\mathrm{91cde}\mid \\ $$$$=\mid\mathrm{9bcde}\mid−\mid\mathrm{9cde}\mid=\mathrm{495}−\mathrm{165}=\mathrm{330} \\ $$$$\left.\mathrm{Brcause}:\mathrm{See}\:\mathrm{the}\:\mathrm{formulas}\:\mathrm{of}\:\mathrm{part2}\right) \\ $$$$\mid\overline {\mathrm{88cde}}\mid=\mid\mathrm{8de}\mid+\mid\mathrm{7de}\mid+\mid\mathrm{6de}\mid+\mid\mathrm{5de}\mid \\ $$$$+\mid\mathrm{4de}\mid+\mid\mathrm{3de}\mid+\mid\mathrm{2de}\mid+\mid\mathrm{1de}\mid=\mathrm{165}−\mathrm{45}=\mathrm{120} \\ $$$$\mid\mathrm{87cde}\mid=\mathrm{120}−\mathrm{36}=\mathrm{84} \\ $$$$\mid\mathrm{86cde}\mid=\mathrm{84}−\mathrm{28}=\mathrm{56} \\ $$$$\mid\mathrm{85cde}\mid=\mathrm{56}−\mathrm{21}=\mathrm{35} \\ $$$$\mid\mathrm{84cde}\mid=\mathrm{35}−\mathrm{15}=\mathrm{20},\mid\mathrm{83cde}\mid=\mathrm{10} \\ $$$$\mid\mathrm{82cde}\mid=\mathrm{4},\mid\mathrm{81cde}\mid=\mathrm{1},\mathrm{so}\:\mathrm{in}\:\mathrm{the}\:\mathrm{total}\:\mathrm{have}\: \\ $$$$\mathrm{120}+\mathrm{84}+\mathrm{56}+\mathrm{35}+\mathrm{20}+\mathrm{10}+\mathrm{5}=\mathrm{330} \\ $$$$\left.\mathrm{c}\right)\mathrm{The}\:\mathrm{case}\:\overline {\mathrm{7bcde}}\:\mathrm{we}\:\mathrm{get}\:\left(\mathrm{see}\:\mathrm{the}\:\mathrm{part3}\right) \\ $$$$\mid\mathrm{7bcde}\mid=\mid\mathrm{8bcde}\mid−\mid\mathrm{8cde}\mid \\ $$$$\mid=\mathrm{330}−\mathrm{120}=\mathrm{210}\:\mathrm{numbers} \\ $$$$\left.\mathrm{d}\right)\mathrm{The}\:\mathrm{case}\:\overline {\mathrm{6bcde}};\mid\mathrm{6bcde}\mid= \\ $$$$\mid\mathrm{7bcde}\mid−\mid\mathrm{7cde}\mid=\mathrm{210}−\mathrm{84}=\mathrm{126} \\ $$$$\left.\mathrm{e}\right)\mathrm{The}\:\mathrm{case}\overline {\mathrm{5bcde}};\mid\:\overline {\mathrm{5bcde}}\mid= \\ $$$$\mid\mathrm{6bcde}\mid−\mid\mathrm{6cde}\mid=\mathrm{126}−\mathrm{56}=\mathrm{70} \\ $$$$\left.\mathrm{f}\right)\mathrm{The}\:\mathrm{case}\:\overline {\mathrm{4bcde}}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mid\mathrm{4bcde}\mid=\mathrm{70}−\mathrm{35}=\mathrm{35}\:\mathrm{numbers} \\ $$$$\left.\mathrm{g}\right)\mathrm{The}\:\mathrm{case}\:\mathrm{3bcde}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mid\mathrm{3bcde}\mid=\mathrm{35}−\mathrm{20}=\mathrm{15}\:\mathrm{numbers} \\ $$$$\mid\mid\mathrm{2bcde}\mid=\mathrm{15}−\mathrm{10}=\mathrm{5},\mid\mathrm{1bcde}\mid=\mathrm{1} \\ $$$$\mathrm{Finaly},\mathrm{in}\:\mathrm{total}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{495}+\mathrm{330}+\mathrm{210}+\mathrm{126}+\mathrm{70}+\mathrm{35}+\mathrm{15}+\mathrm{6}= \\ $$$$\mathrm{1287}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{form}\:\mathrm{abcde}\:\mathrm{satisfy}\: \\ $$$$\mathrm{the}\:\mathrm{condition}\:\mathrm{a}\geqslant\mathrm{b}\geqslant\mathrm{c}\geqslant\mathrm{d}\geqslant\mathrm{e} \\ $$

Commented by prakash jain last updated on 13/Oct/20

2 digits have 54 valid numbers. 10 20 30 40 50 60 70 80 90.

$$\mathrm{2}\:\mathrm{digits}\:\mathrm{have}\:\mathrm{54}\:\mathrm{valid}\:\mathrm{numbers}. \\ $$$$\mathrm{10}\:\mathrm{20}\:\mathrm{30}\:\mathrm{40}\:\mathrm{50}\:\mathrm{60}\:\mathrm{70}\:\mathrm{80}\:\mathrm{90}. \\ $$

Commented by 1549442205PVT last updated on 13/Oct/20

Here i just consider 9 digits 1,2...,9 case 10 digits is similar For 10 digits then ∣ab∣=C_(10) ^2 +10−1 since 00^(−) is rejected formula have given is always true

$$\mathrm{Here}\:\mathrm{i}\:\mathrm{just}\:\mathrm{consider}\:\mathrm{9}\:\mathrm{digits}\:\mathrm{1},\mathrm{2}...,\mathrm{9} \\ $$$$\mathrm{case}\:\mathrm{10}\:\mathrm{digits}\:\mathrm{is}\:\mathrm{similar} \\ $$$$\mathrm{For}\:\mathrm{10}\:\mathrm{digits}\:\mathrm{then} \\ $$$$\mid\mathrm{ab}\mid=\mathrm{C}_{\mathrm{10}} ^{\mathrm{2}} +\mathrm{10}−\mathrm{1}\:\mathrm{since}\:\overline {\mathrm{00}}\:\mathrm{is}\:\mathrm{rejected} \\ $$$$\mathrm{formula}\:\mathrm{have}\:\mathrm{given}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true} \\ $$

Commented by prakash jain last updated on 13/Oct/20

determinant (((x_1 →),9,8,7,6,5,4,3,2,1),((1d),1,1,1,1,1,1,1,1,1),((2d),9,8,7,6,5,4,3,2,1),((3d),(45),(36),(28),(21),(15),(10),6,3,1),((4d),(165),(120),(84),(56),(35),(20),(10),4,1),((5d),(495),(330),(210),(126),(70),(35),(15),5,1),((6d),(1287),(792),(462),(252),(126),(56),(21),6,1)) The very first colums first digit for nd gives the same result as D(n−1) N(n,k)= n digits number with first digit as k N(n,k)=Σ_(j=0) ^k N(n−1,k)

$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\begin{vmatrix}{{x}_{\mathrm{1}} \rightarrow}&{\mathrm{9}}&{\mathrm{8}}&{\mathrm{7}}&{\mathrm{6}}&{\mathrm{5}}&{\mathrm{4}}&{\mathrm{3}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}{d}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{2}{d}}&{\mathrm{9}}&{\mathrm{8}}&{\mathrm{7}}&{\mathrm{6}}&{\mathrm{5}}&{\mathrm{4}}&{\mathrm{3}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{3}{d}}&{\mathrm{45}}&{\mathrm{36}}&{\mathrm{28}}&{\mathrm{21}}&{\mathrm{15}}&{\mathrm{10}}&{\mathrm{6}}&{\mathrm{3}}&{\mathrm{1}}\\{\mathrm{4}{d}}&{\mathrm{165}}&{\mathrm{120}}&{\mathrm{84}}&{\mathrm{56}}&{\mathrm{35}}&{\mathrm{20}}&{\mathrm{10}}&{\mathrm{4}}&{\mathrm{1}}\\{\mathrm{5}{d}}&{\mathrm{495}}&{\mathrm{330}}&{\mathrm{210}}&{\mathrm{126}}&{\mathrm{70}}&{\mathrm{35}}&{\mathrm{15}}&{\mathrm{5}}&{\mathrm{1}}\\{\mathrm{6}{d}}&{\mathrm{1287}}&{\mathrm{792}}&{\mathrm{462}}&{\mathrm{252}}&{\mathrm{126}}&{\mathrm{56}}&{\mathrm{21}}&{\mathrm{6}}&{\mathrm{1}}\end{vmatrix} \\ $$$$\mathrm{The}\:\mathrm{very}\:\mathrm{first}\:\mathrm{colums}\:\mathrm{first}\:\mathrm{digit}\:\mathrm{for}\:{nd}\:\mathrm{gives} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{result}\:\mathrm{as}\:{D}\left({n}−\mathrm{1}\right) \\ $$$${N}\left({n},{k}\right)=\:\mathrm{n}\:\mathrm{digits}\:\mathrm{number}\:\mathrm{with} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{first}\:\mathrm{digit}\:\mathrm{as}\:{k} \\ $$$$\mathrm{N}\left({n},{k}\right)=\underset{{j}=\mathrm{0}} {\overset{{k}} {\sum}}{N}\left({n}−\mathrm{1},{k}\right) \\ $$

Commented by prakash jain last updated on 13/Oct/20

with 9 digits there are 1287 numbers such that x_1 ≥x_2 ≥x_3 ≥x_4 ≥x_5 Σ_(i=0) ^4 ^4 C_i ×^9 C_(i+1) 1×9+4×36+6×84+4×126+1×126= 1287

$$\mathrm{with}\:\mathrm{9}\:\mathrm{digits}\:\mathrm{there}\:\mathrm{are}\:\mathrm{1287}\:\:\mathrm{numbers} \\ $$$$\mathrm{such}\:\mathrm{that}\:{x}_{\mathrm{1}} \geqslant{x}_{\mathrm{2}} \geqslant{x}_{\mathrm{3}} \geqslant{x}_{\mathrm{4}} \geqslant{x}_{\mathrm{5}} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\:^{\mathrm{4}} {C}_{{i}} ×\:^{\mathrm{9}} {C}_{{i}+\mathrm{1}} \\ $$$$\mathrm{1}×\mathrm{9}+\mathrm{4}×\mathrm{36}+\mathrm{6}×\mathrm{84}+\mathrm{4}×\mathrm{126}+\mathrm{1}×\mathrm{126}= \\ $$$$\mathrm{1287} \\ $$

Commented by prakash jain last updated on 13/Oct/20

3 digits ∣abc∣=∣9bc∣+∣8bc∣+∣7bc∣+6bc∣+∣5bc∣ +∣4bc∣+∣3bc∣= i think this missed 222,221,211,111 the total should be 165. Because condition is greater than equal to all above are valid results.

$$\mathrm{3}\:\mathrm{digits} \\ $$$$\mid\mathrm{abc}\mid=\mid\mathrm{9bc}\mid+\mid\mathrm{8bc}\mid+\mid\mathrm{7bc}\mid+\mathrm{6bc}\mid+\mid\mathrm{5bc}\mid \\ $$$$+\mid\mathrm{4bc}\mid+\mid\mathrm{3bc}\mid= \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{this}\:\mathrm{missed} \\ $$$$\mathrm{222},\mathrm{221},\mathrm{211},\mathrm{111} \\ $$$$\mathrm{the}\:\mathrm{total}\:\mathrm{should}\:\mathrm{be}\:\mathrm{165}. \\ $$$$\mathrm{Because}\:\mathrm{condition}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{all}\:\mathrm{above}\:\mathrm{are}\:\mathrm{valid}\:\mathrm{results}. \\ $$

Commented by prakash jain last updated on 13/Oct/20

These are known as binomial cofficients. N(n,k)=OEIS C(n,k−1)

$$\mathrm{These}\:\mathrm{are}\:\mathrm{known}\:\mathrm{as}\:\mathrm{binomial} \\ $$$$\mathrm{cofficients}. \\ $$$${N}\left({n},{k}\right)=\mathrm{OEIS}\:{C}\left({n},{k}−\mathrm{1}\right)\: \\ $$

Commented by prakash jain last updated on 13/Oct/20

http://oeis.org/A000581

Commented by mr W last updated on 13/Oct/20

thank you all!

$${thank}\:{you}\:{all}! \\ $$

Commented by 1549442205PVT last updated on 14/Oct/20

Thank Sir . I missed ∣2de∣=3,∣1de∣=1

$$\mathrm{Thank}\:\mathrm{Sir}\:.\:\mathrm{I}\:\mathrm{missed}\:\mid\mathrm{2de}\mid=\mathrm{3},\mid\mathrm{1de}\mid=\mathrm{1} \\ $$

Answered by mr W last updated on 13/Oct/20

let′s see n digit numbers d_1 d_2 d_3 ...d_n with d_1 ≥d_2 ≥d_3 ≥...≥d_n . we select n_0 times 0, n_1 times 1, n_2 times 2, ..., n_9 times 9 to form such a number. n_0 +n_1 +n_2 +...+n_9 =n ...(i) with 0≤n_i for i=0,1,2,...,9. the number of valid n digit numbers is the number of integer solutions of (i). using generating function (1+x+x^2 +...)^(10) =(1/((1−x)^(10) ))=Σ_(k=0) ^∞ C_9 ^(k+9) x^k the coefficient of x^n term is the answer, which is C_9 ^(n+9) . since the number with n zeros is not valid, we get the final answer C_9 ^(n+9) −1. for 5 digit numbers the answer is C_9 ^(5+9) −1=2001.

$${let}'{s}\:{see}\:{n}\:{digit}\:{numbers}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} ...{d}_{{n}} \\ $$$${with}\:{d}_{\mathrm{1}} \geqslant{d}_{\mathrm{2}} \geqslant{d}_{\mathrm{3}} \geqslant...\geqslant{d}_{{n}} . \\ $$$${we}\:{select}\:{n}_{\mathrm{0}} \:{times}\:\mathrm{0},\:{n}_{\mathrm{1}} \:{times}\:\mathrm{1}, \\ $$$${n}_{\mathrm{2}} \:{times}\:\mathrm{2},\:...,\:{n}_{\mathrm{9}} \:{times}\:\mathrm{9}\:{to}\:{form}\:{such} \\ $$$${a}\:{number}. \\ $$$${n}_{\mathrm{0}} +{n}_{\mathrm{1}} +{n}_{\mathrm{2}} +...+{n}_{\mathrm{9}} ={n}\:\:...\left({i}\right) \\ $$$${with}\:\mathrm{0}\leqslant{n}_{{i}} \:{for}\:{i}=\mathrm{0},\mathrm{1},\mathrm{2},...,\mathrm{9}. \\ $$$${the}\:{number}\:{of}\:{valid}\:{n}\:{digit}\:{numbers} \\ $$$${is}\:{the}\:{number}\:{of}\:{integer}\:{solutions}\:{of} \\ $$$$\left({i}\right). \\ $$$${using}\:{generating}\:{function} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +...\right)^{\mathrm{10}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{10}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{9}} ^{{k}+\mathrm{9}} {x}^{{k}} \\ $$$${the}\:{coefficient}\:{of}\:{x}^{{n}} \:{term}\:{is}\:{the} \\ $$$${answer},\:{which}\:{is}\:{C}_{\mathrm{9}} ^{{n}+\mathrm{9}} .\:{since}\:{the} \\ $$$${number}\:{with}\:{n}\:{zeros}\:{is}\:{not}\:{valid}, \\ $$$${we}\:{get}\:{the}\:{final}\:{answer}\: \\ $$$${C}_{\mathrm{9}} ^{{n}+\mathrm{9}} −\mathrm{1}. \\ $$$${for}\:\mathrm{5}\:{digit}\:{numbers}\:{the}\:{answer}\:{is} \\ $$$${C}_{\mathrm{9}} ^{\mathrm{5}+\mathrm{9}} −\mathrm{1}=\mathrm{2001}. \\ $$

Commented by mr W last updated on 13/Oct/20

exactly sir!

$${exactly}\:{sir}! \\ $$

Commented by mr W last updated on 13/Oct/20

Commented by prakash jain last updated on 13/Oct/20

After looking at your formula, i thought simplifying the result that i got Σ_(j=0) ^(n−1) ^(n−1) C_j ^(10) C_(j+1) =Σ_(j=0) ^(n−1) ^(n−1) C_j ^(10) C_(9−j) =A (1+x)^(9+n) =(1+x)^(n−1) (1+x)^(10) (1+x)^(9+n) =(^(n−1) C_0 +^(n−1) C_1 x+..+^(n−1) C_(n−1) x^(n−1) ) (^(10) C_0 +^(10) C_1 x+^(10) C_2 x^2 +..+^(10) C_(10) x^(10) ) A is same as coeffcient of x^9 in RHS from LHS coefficient of x^9 =^(9+n) C_9

$$\mathrm{After}\:\mathrm{looking}\:\mathrm{at}\:\mathrm{your}\:\mathrm{formula},\:\mathrm{i} \\ $$$$\mathrm{thought}\:\mathrm{simplifying}\:\mathrm{the}\:\mathrm{result}\:\mathrm{that} \\ $$$$\mathrm{i}\:\mathrm{got} \\ $$$$\underset{{j}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\:^{{n}−\mathrm{1}} {C}_{{j}} \:^{\mathrm{10}} {C}_{{j}+\mathrm{1}} \\ $$$$=\underset{{j}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\:^{{n}−\mathrm{1}} {C}_{{j}} \:^{\mathrm{10}} {C}_{\mathrm{9}−{j}} \:=\mathrm{A} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{9}+{n}} =\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} \left(\mathrm{1}+{x}\right)^{\mathrm{10}} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{9}+{n}} =\left(\:^{{n}−\mathrm{1}} {C}_{\mathrm{0}} +\:^{{n}−\mathrm{1}} {C}_{\mathrm{1}} {x}+..+^{{n}−\mathrm{1}} {C}_{{n}−\mathrm{1}} {x}^{{n}−\mathrm{1}} \right) \\ $$$$\:\:\:\:\:\:\:\left(\:^{\mathrm{10}} {C}_{\mathrm{0}} +^{\mathrm{10}} {C}_{\mathrm{1}} {x}+^{\mathrm{10}} {C}_{\mathrm{2}} {x}^{\mathrm{2}} +..+^{\mathrm{10}} {C}_{\mathrm{10}} {x}^{\mathrm{10}} \right) \\ $$$${A}\:\mathrm{is}\:\mathrm{same}\:\mathrm{as}\:\mathrm{coeffcient}\:\mathrm{of}\:{x}^{\mathrm{9}} \:\mathrm{in}\:\mathrm{RHS} \\ $$$$\mathrm{from}\:\mathrm{LHS}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{\mathrm{9}} =\:^{\mathrm{9}+{n}} {C}_{\mathrm{9}} \\ $$

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