A question on permutation & Combinations:

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

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 6We know the direct formula:
a + b + c ... k terms = n.
non negative integral solutions = (n+k1)C(k1)( Note  If you want to learn more on the number of integral solutions kind of problems  refer https://www.mbatious.com/tags/theory of equations )
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 = 17One 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 = 11Now our formula holds good and number of positive solution = (11 + 7  1) C (7  1) = 17C6
We need to remove cases where a, b, c ... f goes beyond 5 (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) = 11C6 ways.Same applicable for b, c, d, e, f also.
So total 6 x 11C6 ways to be removed.Number of ways to get the sum less than or equal to 17 = 17C6 – 6 x 11C6
Hope it is clear.
If you want to strengthen the concept, do the below problems too
 What is the probability of having a sum of 10 after rolling 3 dice?
 Two dice are rolled simultaneously. What is the probability of getting a sum of at least 4?
 When 3 dice are rolled together, probability of the sum being more than 16 is?

thanks for the detailed solution, it's clear now

I have one more query, in the line where you use A = a+6, it is because you want the sum to be less than 6 ? if the sum was 12 instead of 11 in the previous step, it would have been A = a+7. Am I Correct ?

We will solve one more problem to get a better understanding of the concept
Say we need to get a sum of 13 by throwing 3 dice
so a + b + c = 13
Minimum value of a, b and c is 1 so that our equation becomes a + b + c = 13  3 = 10
Number of positive integer solution possible = (10 + 3  1) C (3  1) = 12 C 2 (based on the formula we discussed before)
now let a = A + 6 ( to find the occurrence of values above 6 for a)
A + b + c = 4
Number of positive integer solutions = (4 + 3  1) C (3  1) = 6C2 = 15What does this mean ? There are 15 solutions where a has a value greater than 5 (remember  we consider 0 to 5 as values)
Similarly for b and c we get 15 solutions each where b and c has values greater than 5.
So total 3 x 15 = 45 values need to be removed from the total.
So final answer = 12C2  45
Write down the 15 cases where a + b + c = 10 is true where a > 5 and you will get the concept. (a, b, and c can take values greater than or equal to 0 and less than or equal to 5)

@test1234 Now use the same concept to solve the question given below. Share your steps :)
How many 3 digits numbers have sum of its digit as 15.
Hint  This is just like throwing 3 dice to get a sum 15, where the values can range from 0 to 9 (not 6)
Please share your solution.

a+b+c=15
here, 18, a>9 (invalid case)
put x = x'+9
x'+9 + b + c = 14
x'+b+c = 5
this can be done in 7C2 ways
when b>9, it is an invalid case
put b = y+10
x+y+10+c = 14
x+y+c = 4
this can be done in 6C2 ways
similarly, for cases where c>9, there are 6C2 ways
therefore, total count of 3 digits numbers which have sum of its digit as 15 are:
14C2  (7C2 + 6C2 + 6C2).
Is that right ?

How many 3 digits numbers have sum of its digit as 15.
a + b + c = 15
Number of solutions = ( 15 + 3  1) C (3  1) = 17C2 = 136as a, b and c cannot be more than 10, we need to take it into consideration.
A = a  10
A + b + c = 5
number of solutions = (5 + 3  1) C (3  1) = 7C2 = 21
Considering for b and c, we have to reduce 21 x 3 = 63 from total.136  63 = 73
a cannot be 0 (else it won't be a 3 digit number)
so we need to remove (0, 7, 8); (0, 8, 7); (0, 9, 6) and (0, 6, 9) from the list.So final solution = 73  4 = 69
I will try to collect and share some questions based on this concept as an article which should help.

Sorry, I made a mistake, here is the updated solution,
a+b+c=15
but a>0
So, a`+b+c=14
Number of ways: 16C2
here, when a' is 9, a>9 (invalid case)
put x = a'+9
x'+9 + b + c = 14
x'+b+c = 5
this can be done in 7C2 ways
when b>9, it is an invalid case
put b = y+10
x+y+10+c = 14
x+y+c = 4
this can be done in 6C2 ways
similarly, for cases where c>9, there are 6C2 ways
therefore, total count of 3 digits numbers which have sum of its digit as 15 are:
16C2  (7C2 + 6C2 + 6C2) = 120(21+15+15)
> 12051 = 69
Concept is Clear now, thanks a lot for your help.
(Though I must say, your way of solving is much faster)
reply

All good, happy to help :)