Bunuel wrote:
In how many ways can 5 apples (identical) be distributed among 4 children? (Some children may get no apples.)
(A) 56
(B) 144
(C) 200
(D) 256
(E) 312
TL;DR
Direct Formula:
The total number of ways of dividing n IDENTICAL items among r groups, each one of whom, can receive 0, 1, 2 or more items: (n+r-1)C(r-1)
(5+4-1)C(4-1) = 8C3 = 8*7*6/3! = 8*7 = 56 ways
Intuitive Method 1:
Think of dividers + items (AA|A|A|A)
Anagram Grid: 3 lines & 5 apples
8!/3!5! = 56 ways
Intuitive Method 2:
Think of dividers + items (AA|A|A|A)
# of Dividers = Groups (i.e. people) - 1. = 4 - 1 = 3
Formula: (Items + Dividers)C(Dividers) = (5+3)C(3) = 8C3 = 56 ways
Official Solution
Credit:
Veritas PrepWe have 5 identical apples and 4 children. We want to find the number of ways in which these apples can be distributed among the children.
Method I
5 apples can be distributed in various ways: {5, 0, 0, 0}, {4, 1, 0, 0}, {3, 2, 0, 0}, {3, 1, 1, 0}, {2, 2, 1, 0}, {2, 1, 1, 1}.
{5, 0, 0, 0} means that one child gets all 5 apples and all others get none. Similarly, {4, 1, 0, 0} means that one child gets 4 apples and another child gets 1 apple. No one else gets any apples and so on…
In each one of these cases, various arrangements are possible e.g. take the case of {5, 0, 0, 0}. The first child could get all 5 apples OR the second child could get all 5 apples OR the third child could get all 5 apples OR the fourth child could get all 5 apples. Basically, the number of ways in which these 4 objects – 5, 0, 0 and 0 can be distributed in 4 different spots (i.e. 4 children) is 4!/3! = 4 arrangements (we divide by 3! because three of the objects – 0, 0 and 0 – are identical). This is just our beloved basic counting principle in action.
Similarly, in the case of {4, 1, 0, 0}, we will get 4!/2! = 12 arrangements (since 2 objects are identical) i.e. 5 apples can be distributed among 4 children by giving 4 apples to one child and 1 apple to another child in 12 ways. The first child could get 4 apples and the second child could get 1 apple OR the third child could get 4 apples and the first child could get 1 apple etc.
In the same way, we will get 12 arrangements in each one of these cases: {3, 2, 0, 0}, {3, 1, 1, 0} and {2, 2, 1, 0}. In the case of {2, 1, 1, 1}, we will get 4!/3! = 4 arrangements.
In all, 5 identical apples can be distributed among 4 children in 4 + 12 + 12 + 12 + 12 + 4 = 56 ways
Here, we have just counted out the ways in which 5 things can be distributed in 4 groups. If we miss even one of these cases, all our effort would go waste. Therefore, let’s look at a more analytical method of solving this question.
Method II
Let’s put the 5 apples in a row: A A A A A
We have to split them in 4 groups. The 4 groups will have a one-to-one relation with the 4 children – Apples in the first group will be given to the first child, those in the second group will be given to the second child and so on…
Say we split the apples in 4 groups in the following way: A A l A l A l A
The vertical lines separate one group from the other. The first group has 2 apples and the rest of the three groups have 1 apple each. This means, the first child gets 2 apples and each of the other 3 children get 1 apple each.
The split can also be made in the following way: A l A A l A l A
Here, the second group has 2 apples and the rest of the three groups have 1 apple each. So the second child gets 2 apples and the rest of the 3 children get 1 apple each.
The split can also be made in the following way: l A A A l A l A
Here, the first group has no apples, the second group has 3 apples and the third and the fourth groups have one apple each. The first child gets no apples, the second child gets 3 apples and the other 2 children get 1 apple each.
What I am trying to show here is that arranging these 5 identical As and the 3 identical vertical lines in as many different ways as possible will give us all the ways in which we can distribute 5 identical apples among 4 different children.
In how many different ways can we arrange these 8 objects i.e. 5 identical As and 3 identical vertical lines? In 8!/(5! * 3!) = 56 ways
Answer (A)