There are five different peaches and three different apples. Number of

Shouldn't we consider the case where the first pack has 2 apples and 2 peaches?

Case 1: First pack has 2 Apples and 2 peaches; Second pack has 1 Apple and 3 peaches = 30 ways
Case 2: First pack has 1 Apple and 3 peaches; Second pack has 2 Apples and 2 peaches = 30 ways
So total ways = 60?

Solution:

Since each pack must contain at least one apple, one pack will contain exactly one apple and the other pack will contain exactly two apples. For the pack with one apple, there are three ways of choosing the apple (and the remaining two apples will go in the other pack).

Assuming the apples are distributed, we need to choose three peaches for the pack containing exactly one apple (and the remaining two peaches will go in the other pack). This choice can be made in 5C3 = 5!/(3!*2!) = (5 x 4)/2 = 10 ways.

Since there are 3 ways to distribute the apples and 10 ways to distribute the peaches after the apples are distributed, there are 3 x 10 = 30 ways to divide the fruit.

Alternate Solution:

Let’s calculate the number of ways to divide the fruit into two packs of four without the condition on apples. If the packs were distinguishable, there would have been 8C4 = 8!/(4!*4!) = (8 x 7 x 6 x 5)/(4 x 3 x 2) = 70 ways. However, since the packs are not distinguishable, there are 70/2! = 35 ways to divide the fruit into two packs of four.

Next, notice that the condition “each pack must contain at least one apple” can only fail if all three apples are in one pack. Assuming one of the packs contains all three apples, we have 5 choices for the fourth fruit in that pack (and all the remaining peaches will go in the other pack).

Thus, 5 of the 35 ways to divide the fruit into two packs of four do not meet the criteria and 35 - 5 = 30 of them do.

Answer: D

Because of the conditions placed the question isn’t as bad as it first seems.

(1st) the “stacks” of four fruits are identical groupings. The ordering of the stacks does not matter in the end.

What does matter is which of the different pieces of fruit end up in each of the stacks.

(2nd) since there must be at least 1 apple in each stack, we can only have the apples divided among the 2 identical stacks in one way: [2 ; 1]

How many different ways can we choose 2 DIFFERENT apples to go in stack 1 and 1 apple to go in stack 2?

“3 choose 2” * “1 choose 1” = 3 ways.

AND

(3rd)since the amount of fruit in each pack is already determined (4 and 4) we need to distribute 2 peaches in one stack and 3 peaches in the other stack.

How many different ways can we shuffle around the 5 DIFFERENT peaches such that 2 go in one stack and 3 go in another stack?

“5 choose 2” * “3 choose 3” = 5! / 2! * 3! = 10 ways


(3) * (10) = 30 different ways to distribute

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