Question

An ordinary 6-sided die has a number on each face from 1 to 6 (each number appears on one face). How many ways can I paint two faces of a die red, so that the numbers on the red faces don't add up to 7?

ANSWER Asap

Answer 1

There are 12 ways to paint two faces of a 6-sided die red so that the numbers on the two red faces don't add up to 7, taking into account the pairs that do add up to 7 and avoiding them.

Step-by-step explanation:

To determine the number of ways to paint two faces of a 6-sided die red without the numbers adding up to 7, we first need to consider the pairs of numbers on the die that add up to 7. These pairs are (1, 6), (2, 5), and (3, 4). To avoid these pairs, we cannot paint any of these numbers on the same die red.

Since there are 6 faces on a die, the number of ways to choose 2 faces to paint red is given by the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of items and k is the number of items to be chosen. In this case, n=6 and k=2, so the number of combinations is C(6, 2) = 6! / (2!(6-2)!) = 15.

However, from these 15 combinations, we must subtract the combinations that sum to 7. There are three such combinations, so we subtract 3 from 15, which gives us 12 possible combinations.

Therefore, there are 12 ways to paint two faces of a die red so that the numbers do not add up to 7.