The number of ways in which 8 distinguishable apples can be distribute

Solution:

Let’s first assume that the apples are indistinguishable instead of distinguishable. Then the problem becomes: How many ways can 3 positive integers (none of which is more than 4) add up to 8? We can see that we can have:

1 + 3 + 4 = 8, 2 + 2 + 4 = 8, and 2 + 3 + 3 = 8

For the first addition, there are 3! = 6 way to permute the 3 addends (since they are distinct), while, for each of the last two additions, there are 3! / 2! = 3 ways to permute the 3 addends (since two of them are the same).

Therefore, there are a total of 6 + 3 + 3 = 12 ways, if the 8 apples are indistinguishable. However, since the apples are distinguishable, now let’s consider each of the additions above:

Case 1: 1 + 3 + 4 = 8

Again, this case has a total of 6 additions (besides the one above) that include 3 + 4 + 1 = 8, 4 + 1 + 3 = 8, etc.

Now, let’s look at 1 + 3 + 4 = 8, which means the first boy gets 1 apple, the second 3 apples and the third 4 apples. Since the apples are distinguishable, the number of choices the first boy has is 8, the number of choices the second boy has is 7C3 (after one apple has been chosen by the first boy), and the number of choices the third boy has is 4C4 (after 4 apples have been chosen by the first two boys). Therefore, the number of ways the 8 apples can be distributed is 8 x 7C3 x 4C4 = 8 x (7 x 6 x 5)/(3 x 2) x 1 = 8 x 7 x 5.

Furthermore, regardless of how the 8 apples are distributed to the 3 boys where the distribution (or the partition) of apples is 1, 3, and 4, the number of ways is 8 x 7 x 5. For example, if the first boy gets 3 apples, the second 4 apples and the third 1 apple (i.e., 3 + 4 + 1 = 8), using the same logic above, the number of ways the 8 apples can be distributed is 8C3 x 5C4 x 1C1 =
(8 x 7 x 6)/(3 x 2) x 5 x 1 = 8 x 7 x 5.

Since there are 6 ways we can write 8 as the sum of 1, 3, and 4 where the order of the addends matters (i.e., 6 partitions of 8 apples into 3 piles consisting 1, 3, and 4 apples), we see that the total number of ways in case 1 is 6 x (8 x 7 x 5) = 8 x (7 x 6 x 5) = 8 x 7P3.
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Case 2: 2 + 2 + 4 = 8

Again, this case has a total of 3 additions (besides the one above) that include 2 + 4 + 2 = 8 and 4 + 2 + 2 = 8.

Using the same logic as in case 1, we see that if the first boy gets 2 apples, the second 2 apples and the third 4 apples (i.e., 2 + 2 + 4 = 8), the number of ways the 8 apples can be distributed is 8C2 x 6C2 x 4C4 = (8 x 7)/2 x (6 x 5)/2 x 1 = 4 x 7 x 3 x 5. Since there are 3 ways we can write 8 as the sum of 2, 2, and 4 where the order of the addends matters (i.e., 3 partitions of 8 apples into 3 piles consisting 2, 2, and 4 apples), we see that the total number of ways in case 2 is
3 x (4 x 7 x 3 x 5) = 36 x 7 x 5 = 6 x (7 x 6 x 5) = 6 x 7P3.

Case 3: 2 + 3 + 3 = 8

Using the same argument as in case 2, there will be also 6 x 7P3 ways the 8 apples can be partitioned into 3 piles consisting of 2, 3, and 3 apples.

Therefore, the total number of ways in all 3 cases is 8 x 7P3 + 6 x 7P3 + 6 x 7P3 = (8 + 6 + 6) x 7P3 = 22 x 7P3. Therefore, K is 22.

Answer: D