Category: H1 Math

Nine gifts, three of which are identical, and the rest are distinct, are distributed among five people without restrictions on the number of gifts a person can have. By first considering the number of ways to distribute the distinct gifts or otherwise, find the number of ways that the nine gifts can be distributed.

First, consider the number of ways to distribute the 6 distinct gifts:

Gift A can be given to either Person P, Q, R, S or T: 5 ways

Gift B can be given to either Person P, Q, R, S or T: 5 ways

Gift C: 5 ways, Gift D: 5 ways, Gift E: 5 ways, Gift F: 5 ways

So, for 6 distinct gifts, number of ways to distribute to 5 persons = 5⁶

Next, consider the number of ways to distribute the 3 identical gifts:

Case 1: 3 of the persons get 1 of these gifts each = 5C3 = 10

Case 2: 1 of the persons gets 1 of these gifts, another person gets 2 of these gifts:

Case 3: 1 of the persons get all 3 of these gifts = 5C1 = 5

Total number of ways = (10 + 20 + 5) x 5⁶

                                    = 546 875

Consider the factors of 2^6. There are 7 factors: 1, 2, 2^2, 2^3, ……, 2^6.
Similarly, for 3 there are 2 factors: 1, 3; and
similarly, for 643, there are 2 factors: 1, 643
So the total number of factors will be 7 x 2 x 2 – 2 (1 x 1 x 1 = 1, and 2^6 x 3 x 643 = 123456 need to be minus off) = 26 positive factors