jmc

Question

I have a bag with 6 marbles numbered from 1 to 6. Mathew has a bag with 12 marbles numbered from 1 to 12. Mathew chooses one marble from his bag and I choose two from mine. In how many ways can we choose the marbles (where the order of my choices does matter) such that the sum of the numbers on my marbles equals the number on his?

Accepted Answer

We could proceed by listing the various cases, depending on which number Mathew draws.

Mathew’s number 123456789101112 My pair of numbers - - (1, 2), (2, 1) (1, 3), (3, 1) (1, 4), (2, 3), (3, 2), (4, 1) (1, 5), (2, 4), (4, 2), (5, 1) (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) (2, 6), (3, 5), (5, 3), (6, 2) (3, 6), (4, 5), (5, 4), (6, 3) (4, 6), (6, 4) (5, 6), (6, 5) -

The answer is

2 + 2 + 4 + 4 + 6 + 4 + 4 + 2 + 2 = 30 .

There is a much easier solution: there are

6 \times 5 = 30

choices for the two marbles that I draw. Once I draw my two marbles, there is only

1

way for Mathew to draw the marble which has the sum of my two marbles. So the total number of possibilities is just equal to the number of ways in which I can draw my two marbles, which is

30 .