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

H o w m a n y f i v e d i g i t n u m b e r s

e x i s t s u c h t h a t t h e s u m o f t h e i r

d i g i t s e q u a l s 43 ?

H o w m a n y e x i s t i f t h e s u m i s

39 ?

Commented by 123456 last updated on 08/Jul/15

a + b + c + d + e = k

a \in {1, 2, 3, \dots, 9}

b, c, d, e \in {0, 1, \dots, 9}

k \in {39, 43}

S_{1} \to k = 39

S_{2} \to k = 43

Answered by prakash jain last updated on 09/Jul/15

Five digits all 9 s = 45

To get 43 we need to subtract 2 .

00011 can be subtracted in \frac{5!}{3! 2!} ways .

00002 can be subtracted in \frac{5!}{4! 1!} ways .

Total ways to form 5 digit number with

sum of digits 43 = \frac{5!}{3! 2!} + \frac{5!}{4! 1!} = 10 + 5 = 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

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 1 s and 1 2 (11112) = \frac{5!}{4! 1!} = 5

3 1 s, 1 0 and 1 3 (11103) = \frac{5!}{3! 1! 1!} = 20

2 1 s, 1 0 and 2 2 s (11022) = \frac{5!}{2! 2! 1!} = 30

2 1 s, 2 0 s and 1 4 (11004) = \frac{5!}{2! 2! 1!} = 30

1 1 s, 2 0 s and 1 2, 1 3 (10023) = \frac{5!}{2! 1! 1! 1!} = 60

1 1 s, 3 0 s and 1 5 (10005) = \frac{5!}{1! 3! 1!} = 20

0 1 s, 3 2 s and 2 0 s (22200) = \frac{5!}{2! 3!} = 10

0 1 s, 1 2, 3 0 and 1 4 (20004) = \frac{5!}{3!} = 20

0 1 s, 0 2 s, 2 3 s (00033) = \frac{5!}{3! 2!} = 10

0 1 s, 0 2 s, 1 6 (00006) = \frac{5!}{1! 4!} = 5

Total = sum of all above .

Commented by 112358 last updated on 09/Jul/15

T h a n k s

Commented by 112358 last updated on 09/Jul/15

T h e r e a r e a c o u p l e c o m p u t a t i o n

e r r o r s b u t I u n d e r s t a n d t h e

s o l u t i o n .

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