Question

Find the number of ways in which we can get a sum less than or equal to 17 by throwing six distinct dices.

• A. 9604
• B. 6832
• C. 8500
• D. 5748

Answer 1

The correct option is A 9604

We are asked to find the number of ways in which 6 dice can give a sum less than or equal to 17 means, a + b + c + d + e + f < = 17
a, b, c ... f can take values from 1 to 6
We know the direct formula:
a + b + c ... k terms = n.
non negative integral solutions = ${}^{(n+k-1)}{C}_{(k-1)}$
But here is a catch. This holds for "Equal to n" and not for "Less than or Equal to n".
To solve this, add a dummy variable. (say, g)
so we have a + b + c + d + e + f + g = 17
One important thing here is than minimum value of a, b, c ... f is 1 ( not 0) so our equation reduces to
a + b + c + d + e + f + g = 17 - 6 = 11
Now our formula holds good and number of positive solutions = ${}^{(11+7-1)}{C}_{(7-1)}{=}^{17}{C}_6$
We need to remove cases where a, b, c ... f goes beyond 6 (as then then the value goes beyond 6 which is not possible with a dice)
Take a = A + 6
So the equation becomes A + b + c + d + e + f + g = 11 - 6 = 5
Number of non-negative integral solutions = ${}^{(5+7-1)}{C}_{(7-1)}={}^{11}{C}_6$ ways.
Same applicable for b, c, d, e, f also.
So total $6\times {}^{11}{C}_6$ ways to be removed.
Number of ways to get the sum less than or equal to $17={}^{17}{C}_6-(6\times {}^{11}{C}_6)=9604$.