Menu Close

How-many-five-digit-numbers-exist-such-that-the-sum-of-their-digits-equals-43-How-many-exist-if-the-sum-is-39-




Question Number 1164 by 112358 last updated on 08/Jul/15
How many five digit numbers  exist such that the sum of their  digits equals 43?   How many exist if the sum is  39?
$${How}\:{many}\:{five}\:{digit}\:{numbers} \\ $$$${exist}\:{such}\:{that}\:{the}\:{sum}\:{of}\:{their} \\ $$$${digits}\:{equals}\:\mathrm{43}?\: \\ $$$${How}\:{many}\:{exist}\:{if}\:{the}\:{sum}\:{is} \\ $$$$\mathrm{39}? \\ $$$$ \\ $$
Commented by 123456 last updated on 08/Jul/15
a+b+c+d+e=k  a∈{1,2,3,...,9}  b,c,d,e∈{0,1,...,9}  k∈{39,43}  S_1 →k=39  S_2 →k=43
$${a}+{b}+{c}+{d}+{e}={k} \\ $$$${a}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},…,\mathrm{9}\right\} \\ $$$${b},{c},{d},{e}\in\left\{\mathrm{0},\mathrm{1},…,\mathrm{9}\right\} \\ $$$${k}\in\left\{\mathrm{39},\mathrm{43}\right\} \\ $$$$\mathrm{S}_{\mathrm{1}} \rightarrow{k}=\mathrm{39} \\ $$$$\mathrm{S}_{\mathrm{2}} \rightarrow{k}=\mathrm{43} \\ $$
Answered by prakash jain last updated on 09/Jul/15
Five digits all 9s=45  To get 43 we need to subtract 2.  00011 can be subtracted in ((5!)/(3!2!)) ways.  00002 can be subtracted in ((5!)/(4!1!)) ways.  Total ways to form 5 digit number with  sum of digits 43=((5!)/(3!2!))+((5!)/(4!1!))=10+5=15
$$\mathrm{Five}\:\mathrm{digits}\:\mathrm{all}\:\mathrm{9s}=\mathrm{45} \\ $$$$\mathrm{To}\:\mathrm{get}\:\mathrm{43}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{subtract}\:\mathrm{2}. \\ $$$$\mathrm{00011}\:\mathrm{can}\:\mathrm{be}\:\mathrm{subtracted}\:\mathrm{in}\:\frac{\mathrm{5}!}{\mathrm{3}!\mathrm{2}!}\:\mathrm{ways}. \\ $$$$\mathrm{00002}\:\mathrm{can}\:\mathrm{be}\:\mathrm{subtracted}\:\mathrm{in}\:\frac{\mathrm{5}!}{\mathrm{4}!\mathrm{1}!}\:\mathrm{ways}. \\ $$$$\mathrm{Total}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{form}\:\mathrm{5}\:\mathrm{digit}\:\mathrm{number}\:\mathrm{with} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{digits}\:\mathrm{43}=\frac{\mathrm{5}!}{\mathrm{3}!\mathrm{2}!}+\frac{\mathrm{5}!}{\mathrm{4}!\mathrm{1}!}=\mathrm{10}+\mathrm{5}=\mathrm{15} \\ $$
Commented by prakash jain last updated on 09/Jul/15
88999  89899   89989   89998  98899  98989   98998  99889  99898  99988  79999  97999   99799   99979  99997
$$\mathrm{88999}\:\:\mathrm{89899}\:\:\:\mathrm{89989}\:\:\:\mathrm{89998} \\ $$$$\mathrm{98899}\:\:\mathrm{98989}\:\:\:\mathrm{98998} \\ $$$$\mathrm{99889}\:\:\mathrm{99898} \\ $$$$\mathrm{99988} \\ $$$$\mathrm{79999}\:\:\mathrm{97999}\:\:\:\mathrm{99799}\:\:\:\mathrm{99979}\:\:\mathrm{99997} \\ $$
Answered by prakash jain last updated on 09/Jul/15
To get sum 39 we need to subtract 6.  6 can be obtained by  4 1s and 1 2(11112)=((5!)/(4!1!))=5  3 1s, 1 0 and 1 3(11103)=((5!)/(3!1!1!))=20  2 1s, 1 0 and 2 2s(11022)=((5!)/(2!2!1!))=30  2 1s, 2 0s and 1 4(11004)=((5!)/(2!2!1!))=30  1 1s, 2 0s and 1 2, 1 3(10023)=((5!)/(2!1!1!1!))=60  1 1s, 3 0s and 1 5(10005)=((5!)/(1!3!1!))=20  0 1s, 3 2s and 2 0s(22200)=((5!)/(2!3!))=10  0 1s, 1 2 , 3 0and 1 4(20004)=((5!)/(3!))=20  0 1s, 0 2s , 2 3s (00033)=((5!)/(3!2!))=10  0 1s, 0 2s , 1 6 (00006)=((5!)/(1!4!))=5  Total=sum of all above.
$$\mathrm{To}\:\mathrm{get}\:\mathrm{sum}\:\mathrm{39}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{subtract}\:\mathrm{6}. \\ $$$$\mathrm{6}\:\mathrm{can}\:\mathrm{be}\:\mathrm{obtained}\:\mathrm{by} \\ $$$$\mathrm{4}\:\mathrm{1s}\:\mathrm{and}\:\mathrm{1}\:\mathrm{2}\left(\mathrm{11112}\right)=\frac{\mathrm{5}!}{\mathrm{4}!\mathrm{1}!}=\mathrm{5} \\ $$$$\mathrm{3}\:\mathrm{1s},\:\mathrm{1}\:\mathrm{0}\:\mathrm{and}\:\mathrm{1}\:\mathrm{3}\left(\mathrm{11103}\right)=\frac{\mathrm{5}!}{\mathrm{3}!\mathrm{1}!\mathrm{1}!}=\mathrm{20} \\ $$$$\mathrm{2}\:\mathrm{1s},\:\mathrm{1}\:\mathrm{0}\:\mathrm{and}\:\mathrm{2}\:\mathrm{2s}\left(\mathrm{11022}\right)=\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{2}!\mathrm{1}!}=\mathrm{30} \\ $$$$\mathrm{2}\:\mathrm{1s},\:\mathrm{2}\:\mathrm{0s}\:\mathrm{and}\:\mathrm{1}\:\mathrm{4}\left(\mathrm{11004}\right)=\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{2}!\mathrm{1}!}=\mathrm{30} \\ $$$$\mathrm{1}\:\mathrm{1s},\:\mathrm{2}\:\mathrm{0s}\:\mathrm{and}\:\mathrm{1}\:\mathrm{2},\:\mathrm{1}\:\mathrm{3}\left(\mathrm{10023}\right)=\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{1}!\mathrm{1}!\mathrm{1}!}=\mathrm{60} \\ $$$$\mathrm{1}\:\mathrm{1s},\:\mathrm{3}\:\mathrm{0s}\:\mathrm{and}\:\mathrm{1}\:\mathrm{5}\left(\mathrm{10005}\right)=\frac{\mathrm{5}!}{\mathrm{1}!\mathrm{3}!\mathrm{1}!}=\mathrm{20} \\ $$$$\mathrm{0}\:\mathrm{1s},\:\mathrm{3}\:\mathrm{2s}\:\mathrm{and}\:\mathrm{2}\:\mathrm{0s}\left(\mathrm{22200}\right)=\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{3}!}=\mathrm{10} \\ $$$$\mathrm{0}\:\mathrm{1s},\:\mathrm{1}\:\mathrm{2}\:,\:\mathrm{3}\:\mathrm{0and}\:\mathrm{1}\:\mathrm{4}\left(\mathrm{20004}\right)=\frac{\mathrm{5}!}{\mathrm{3}!}=\mathrm{20} \\ $$$$\mathrm{0}\:\mathrm{1s},\:\mathrm{0}\:\mathrm{2s}\:,\:\mathrm{2}\:\mathrm{3s}\:\left(\mathrm{00033}\right)=\frac{\mathrm{5}!}{\mathrm{3}!\mathrm{2}!}=\mathrm{10} \\ $$$$\mathrm{0}\:\mathrm{1s},\:\mathrm{0}\:\mathrm{2s}\:,\:\mathrm{1}\:\mathrm{6}\:\left(\mathrm{00006}\right)=\frac{\mathrm{5}!}{\mathrm{1}!\mathrm{4}!}=\mathrm{5} \\ $$$$\mathrm{Total}=\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{above}. \\ $$$$ \\ $$
Commented by 112358 last updated on 09/Jul/15
Thanks
$${Thanks} \\ $$
Commented by 112358 last updated on 09/Jul/15
There are a couple computation   errors but I understand the   solution.
$${There}\:{are}\:{a}\:{couple}\:{computation}\: \\ $$$${errors}\:{but}\:{I}\:{understand}\:{the}\: \\ $$$${solution}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *