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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.

$${There}\:{are}\:{a}\:{couple}\:{computation}\: \\ $$$${errors}\:{but}\:{I}\:{understand}\:{the}\: \\ $$$${solution}. \\ $$

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