@Naman-Jain-0

Number of ways to get a sum greater than 17 = Total - Number of ways to get a sum lesser than or equal to 17

6 dice and each dice can give 6 options. Total cases = 6^6

Now we will find the number of ways to get a sum lesser than or equal to 17

a + b + c + d + e + f ≤ 17

a , b, c ... f can take values from 1 to 6

Because of ≤, add a dummy variable. (say, g)

so we have a + b + c + d + e + f + g = 17

Now we will discuss a very important point. a, b, c .. f are all natural numbers and for g - 0 is possible. Means this is neither in case 1 or case 2. To resolve this let subtract 1 from a, b, c .. f so that they can be given a value of 0 also. Our equation becomes

a + b + c + d + e + f +g = 17 - 6 = 11

Concept : a + b + c ... (k terms) = n has (n + k - 1) C (k - 1) non negative integer solutions

Number of ways = (11 + 7 - 1) C (7 - 1) = 17C6

now we have to remove cases where a > 6.

say a = A + 6

So, A + b + c + d + e + f + g = 11 - 6 = 5

Number of ways which a can be higher than 6 (invalid cases for a) = 11C6 ways.

6 x 11C6 invalid cases if we include for b, c, d, e and f also.

So Final - 17C6 - 6 x 11C6

Now coming back to the original question,

Number of ways to get a sum greater than 17 = Total - Number of ways to get a sum lesser than or equal to 17

= 6^6 - (17C6 - 6 x 11C6)

If you need a clear understanding of this concept, refer the below article. Happy Learning!

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Solving Combinatorics Problems Using Stars & Bars Concept