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Question Number 117061 by ZiYangLee last updated on 09/Oct/20
How many 5-digits positive integers  x_1 x_2 x_3 x_4 x_5  are there such that   x_1 ≤x_2 ≤x_3 ≤x_4 ≤x_5 ?
$$\mathrm{How}\:\mathrm{many}\:\mathrm{5}-\mathrm{digits}\:\mathrm{positive}\:\mathrm{integers} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} {x}_{\mathrm{4}} {x}_{\mathrm{5}} \:\mathrm{are}\:\mathrm{there}\:\mathrm{such}\:\mathrm{that}\: \\ $$$${x}_{\mathrm{1}} \leqslant{x}_{\mathrm{2}} \leqslant{x}_{\mathrm{3}} \leqslant{x}_{\mathrm{4}} \leqslant{x}_{\mathrm{5}} ? \\ $$
Commented by ZiYangLee last updated on 09/Oct/20
nope is 715 instead
$${nope}\:{is}\:\mathrm{715}\:{instead} \\ $$
Commented by MJS_new last updated on 09/Oct/20
2 digits n=Σ_(a=1) ^9 a=45  3 digits n=Σ_(b=1) ^9 (Σ_(a=1) ^b a)=165  4 digits n=Σ_(c=1) ^9 (Σ_(b=1) ^c (Σ_(a=1) ^b a))=495  5 digits n=Σ_(d=1) ^9 (Σ_(c=1) ^d (Σ_(b=1) ^c (Σ_(a=1) ^b a)))=1287  k digits n=(1/(k!))Π_(j=0) ^(k−1) (j+9)=(((k+8)!)/(8!k!))
$$\mathrm{2}\:\mathrm{digits}\:{n}=\underset{{a}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}{a}=\mathrm{45} \\ $$$$\mathrm{3}\:\mathrm{digits}\:{n}=\underset{{b}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\left(\underset{{a}=\mathrm{1}} {\overset{{b}} {\sum}}{a}\right)=\mathrm{165} \\ $$$$\mathrm{4}\:\mathrm{digits}\:{n}=\underset{{c}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\left(\underset{{b}=\mathrm{1}} {\overset{{c}} {\sum}}\left(\underset{{a}=\mathrm{1}} {\overset{{b}} {\sum}}{a}\right)\right)=\mathrm{495} \\ $$$$\mathrm{5}\:\mathrm{digits}\:{n}=\underset{{d}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\left(\underset{{c}=\mathrm{1}} {\overset{{d}} {\sum}}\left(\underset{{b}=\mathrm{1}} {\overset{{c}} {\sum}}\left(\underset{{a}=\mathrm{1}} {\overset{{b}} {\sum}}{a}\right)\right)\right)=\mathrm{1287} \\ $$$${k}\:\mathrm{digits}\:{n}=\frac{\mathrm{1}}{{k}!}\underset{{j}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left({j}+\mathrm{9}\right)=\frac{\left({k}+\mathrm{8}\right)!}{\mathrm{8}!{k}!} \\ $$
Commented by MJS_new last updated on 09/Oct/20
how you get 715?
$$\mathrm{how}\:\mathrm{you}\:\mathrm{get}\:\mathrm{715}? \\ $$
Answered by Olaf last updated on 09/Oct/20
A number is a plaindrome in a given  base b (often 10 or 16) if its digits are  in nondecreasing order in that base.  A plaindrome in which the digits are  strictly increasing is called metadrome,  while numbers whose digits are   nonincreasing and strictly decreasing  are called nialpdromes and katadromes.    See OEIS, Sequence A009994 (base 10)  See OEIS, Sequence A023757 (base 16)    See numbersaplenty.com for calculous.  (in english)    See villemin.gerard.free.fr  (but this is in french)
$$\mathrm{A}\:\mathrm{number}\:\mathrm{is}\:\mathrm{a}\:\mathrm{plaindrome}\:\mathrm{in}\:\mathrm{a}\:\mathrm{given} \\ $$$$\mathrm{base}\:{b}\:\left(\mathrm{often}\:\mathrm{10}\:\mathrm{or}\:\mathrm{16}\right)\:\mathrm{if}\:\mathrm{its}\:\mathrm{digits}\:\mathrm{are} \\ $$$$\mathrm{in}\:\mathrm{nondecreasing}\:\mathrm{order}\:\mathrm{in}\:\mathrm{that}\:\mathrm{base}. \\ $$$$\mathrm{A}\:\mathrm{plaindrome}\:\mathrm{in}\:\mathrm{which}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{are} \\ $$$$\mathrm{strictly}\:\mathrm{increasing}\:\mathrm{is}\:\mathrm{called}\:\mathrm{metadrome}, \\ $$$$\mathrm{while}\:\mathrm{numbers}\:\mathrm{whose}\:\mathrm{digits}\:\mathrm{are}\: \\ $$$$\mathrm{nonincreasing}\:\mathrm{and}\:\mathrm{strictly}\:\mathrm{decreasing} \\ $$$$\mathrm{are}\:\mathrm{called}\:\mathrm{nialpdromes}\:\mathrm{and}\:\mathrm{katadromes}. \\ $$$$ \\ $$$$\mathrm{See}\:\mathrm{OEIS},\:\mathrm{Sequence}\:\mathrm{A009994}\:\left(\mathrm{base}\:\mathrm{10}\right) \\ $$$$\mathrm{See}\:\mathrm{OEIS},\:\mathrm{Sequence}\:\mathrm{A023757}\:\left(\mathrm{base}\:\mathrm{16}\right) \\ $$$$ \\ $$$$\mathrm{See}\:\mathrm{numbersaplenty}.\mathrm{com}\:\mathrm{for}\:\mathrm{calculous}. \\ $$$$\left(\mathrm{in}\:\mathrm{english}\right) \\ $$$$ \\ $$$$\mathrm{See}\:\mathrm{villemin}.\mathrm{gerard}.\mathrm{free}.\mathrm{fr} \\ $$$$\left(\mathrm{but}\:\mathrm{this}\:\mathrm{is}\:\mathrm{in}\:\mathrm{french}\right) \\ $$$$ \\ $$
Answered by mr W last updated on 11/Oct/20
we see in such a 5 digit number the  digit zero is not allowed. so we can  only select from digits 1 to 9.  say the number of digit 1 is n_1 , the  number of digit 2 is n_2 , etc.  n_1 +n_2 +n_3 +...+n_9 =5   ...(i)  with 0≤n_1 ,n_2 ,...,n_9 ≤5  we know that we can always arrange  the n_1  times 1, n_2  times 2, ..., n_9  times  9 in increasing order to form a 5  digit number x_1 x_2 ...x_5  such that  x_1 ≤x_2 ≤...≤x_5 .  so the number of non−negative  integer solutions of eqn. (i) is the  answer of our question.  the number of non−negative   integer solutions of eqn. (i) is the  coefficient of term x^5  of the following  generating function  (1+x+x^2 +x^3 +...)^9 =(1/((1−x)^9 ))=Σ_(k=0) ^∞ C_8 ^(k+8) x^k   the cof. of x^5  term is C_8 ^(13) =1287.    generally the number of k digit  numbers is C_8 ^(k+8) =(((k+8)!)/(8!k!)).
$${we}\:{see}\:{in}\:{such}\:{a}\:\mathrm{5}\:{digit}\:{number}\:{the} \\ $$$${digit}\:{zero}\:{is}\:{not}\:{allowed}.\:{so}\:{we}\:{can} \\ $$$${only}\:{select}\:{from}\:{digits}\:\mathrm{1}\:{to}\:\mathrm{9}. \\ $$$${say}\:{the}\:{number}\:{of}\:{digit}\:\mathrm{1}\:{is}\:{n}_{\mathrm{1}} ,\:{the} \\ $$$${number}\:{of}\:{digit}\:\mathrm{2}\:{is}\:{n}_{\mathrm{2}} ,\:{etc}. \\ $$$${n}_{\mathrm{1}} +{n}_{\mathrm{2}} +{n}_{\mathrm{3}} +…+{n}_{\mathrm{9}} =\mathrm{5}\:\:\:…\left({i}\right) \\ $$$${with}\:\mathrm{0}\leqslant{n}_{\mathrm{1}} ,{n}_{\mathrm{2}} ,…,{n}_{\mathrm{9}} \leqslant\mathrm{5} \\ $$$${we}\:{know}\:{that}\:{we}\:{can}\:{always}\:{arrange} \\ $$$${the}\:{n}_{\mathrm{1}} \:{times}\:\mathrm{1},\:{n}_{\mathrm{2}} \:{times}\:\mathrm{2},\:…,\:{n}_{\mathrm{9}} \:{times} \\ $$$$\mathrm{9}\:{in}\:{increasing}\:{order}\:{to}\:{form}\:{a}\:\mathrm{5} \\ $$$${digit}\:{number}\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} …{x}_{\mathrm{5}} \:{such}\:{that} \\ $$$${x}_{\mathrm{1}} \leqslant{x}_{\mathrm{2}} \leqslant…\leqslant{x}_{\mathrm{5}} . \\ $$$${so}\:{the}\:{number}\:{of}\:{non}−{negative} \\ $$$${integer}\:{solutions}\:{of}\:{eqn}.\:\left({i}\right)\:{is}\:{the} \\ $$$${answer}\:{of}\:{our}\:{question}. \\ $$$${the}\:{number}\:{of}\:{non}−{negative}\: \\ $$$${integer}\:{solutions}\:{of}\:{eqn}.\:\left({i}\right)\:{is}\:{the} \\ $$$${coefficient}\:{of}\:{term}\:{x}^{\mathrm{5}} \:{of}\:{the}\:{following} \\ $$$${generating}\:{function} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{\mathrm{9}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{9}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{8}} ^{{k}+\mathrm{8}} {x}^{{k}} \\ $$$${the}\:{cof}.\:{of}\:{x}^{\mathrm{5}} \:{term}\:{is}\:{C}_{\mathrm{8}} ^{\mathrm{13}} =\mathrm{1287}. \\ $$$$ \\ $$$${generally}\:{the}\:{number}\:{of}\:{k}\:{digit} \\ $$$${numbers}\:{is}\:{C}_{\mathrm{8}} ^{{k}+\mathrm{8}} =\frac{\left({k}+\mathrm{8}\right)!}{\mathrm{8}!{k}!}. \\ $$
Commented by mr W last updated on 11/Oct/20

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