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.



  • @test1234

    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)

    ( 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 = 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 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

    1. What is the probability of having a sum of 10 after rolling 3 dice?
    2. Two dice are rolled simultaneously. What is the probability of getting a sum of at least 4?
    3. 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 ?



  • @test1234

    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 = 15

    What 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 ?



  • @test1234

    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 = 136

    as 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)
    -> 120-51 = 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 :)


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