Question

In: Statistics and Probability

On the table is a box of donuts containing 8 donuts, each of a different kind....

On the table is a box of donuts containing 8 donuts, each of a different kind.
a) You have first dibs on the donuts, and you can grab as many as you like or forgo them entirely. In how different many ways can you take the donuts home with you?

b) Suppose there is a condition that you must take at least one but no more than five. In how many ways can you take the donut(s) with you?

c) Suppose all 8 donuts are to be distributed among three people, with no restriction on the
number of donuts each person can have. (This includes the possibility that someone ends
up empty-handed.) In how many ways can this be done?

Note: Each donut is different, so that the scenario where Person 1 gets Donut A and
Donut B while the other 6 donuts go to Person 2 (and thus Person 3 gets nothing) is
distinct from the one where Person 1 gets Donut C and Donut F and the other 6 donuts go
to Person 2.

Expert Solution

The box has 8 donuts, each of a different kind.

a)You can take 0,1,2,...,or 8 donuts with you.

The required sum is

b) Here you must take at least one but no more than five.

The required sum is

c)The number of ways of arranging identical balls in distinct urns is the number of terms in the expansion of . This is .

Here taking , the required number of ways is

orchestra answered 2 years ago

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