# person gives n=7 gifts to 6 friends. friends are picked randomly. Find the probability that at least 1 friend does not receive a gift.?

### 1 Answer

- PuzzlingLv 71 month ago
Let's first count the total way that we can distribute the gifts.

There are 6 choices of recipient for each of the 7 gifts (repeats allowed obviously).

6^7 = 279,936 ways

Now rather than looking for the ways to hand out the gifts so that at least one person *doesn't* get a gift, let's count the ways that everyone can get at least one gift.

If we hand out the gifts, one person will get 2 gifts and the rest will get 1.

First pick the person that gets the two gifts:

6C1 = 6 ways

Then pick the 2 gifts that person gets:

7C2 = (7 * 6) / 2 = 21 ways

Finally there are 5! = 120 ways to hand out the remaining gifts to the other 5 friends.

6 * 21 * 120 = 15,120 ways to hand out gifts so that EVERYONE has at least one gift.

Subtract and you see that there are 264,816 ways to hand out the gifts so that some people don't get gifts.

Now divide that by the total ways to distribute gifts (279,936) to get the probability.

Answer:

264816/279936

≈ 0.94598765432

About 94.6%