@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 17a + b + c + d + e + f ≤ 17a , 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 + 6So, A + b + c + d + e + f + g = 11 - 6 = 5Number 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!(adsbygoogle = window.adsbygoogle || []).push({});
Solving Combinatorics Problems Using Stars & Bars Concept