Question

how many 6 digit numbers are possible with the property that the sum of their digits is 5

Answer 1

Let 6 digits numbers be ABCDEF.

Given that their sum is 5.

Here, 

n = 5, r = 6.

Apply the formula:

= > (n + r - 1)c (r - 1)

= > (5 + 6 - 1)c(6 - 1)

= > (10c5)

= > 10!/5!(10 - 5!)

= > 10!/5!5!

= > 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 *  2 * 1/5 * 4 * 3 * 2 * 1 * 5 * 4 * 3 * 2 * 1

= > 252.



Therefore, the possible 6 -digit numbers are 252.


Hope it helps!

Answer 2

There are 126 such numbers.

Let us assume the six digit number to be of the form A B C D E F.

For six digit numbers to be formed, A can never be zero. Thus, we must ensure that zero comes only in the place of either B, C, D, E or F.

Now, for the sum of the digits to be 5, A + B + C + D + E + F = 5.

This can be achieved in the following ways:

(A,B,C,D,E,F) = (1,0,0,0,0,4) -> 5 ways

= (1,1,1,1,1,0) -> 5 ways

= (1,0,0,0,1,3) -> 20 ways

= (1,0,0,0,2,2) -> 10 ways

= (1,0,0,1,1,2) -> 30 ways

= (2,0,0,0,0,3) -> 5 ways

= (2,0,0,0,1,2) -> 20 ways

= (2,0,0,1,1,1) -> 10 ways

= (3,0,0,0,0,2) -> 5 ways

= (3,0,0,0,1,1) -> 10 ways

= (4,0,0,0,0,1) -> 5 ways

= (5,0,0,0,0,0) -> 1 way

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126 ways

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