In how many ways can 5 apples (identical) be distributed among 4 child

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 Prep

We 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)

Yes, the above formula includes zero.
The number of ways of distributing 'n' identical objects amongst 'r' groups such that each group gets at least 1 object is given by

\({n-1}_C_{r-1}\)

The number of ways of distributing 20 apples among 4 baskets such that each basket has at least 1 apple is

\({20-1}_C_{4-1}\) = \({19}_C_{3}\)

lets try to solve it differently
if you have 5 apples A,A,A,A,A and you have to put three partition +,+,+

it can be AAA+AA+nothing+nothing
or
A+A+AA+A
or
nothing+AA+AAA+nothing

ultimately you are arranging 5 A and 3+

question becomes In how many different way you can arrange AAAAA+++

\(= \frac{8!}{(3!*5!)}\)=56

The answer is A.
Here is how I found the answer
Suppose we have five letter A’s representing apples. Let us also use three *’s which will represent partitions (four) between the apples belonging to different children. We order the A’s and *’s as we like and interpret all A’s before the first * as being apples belonging to Kathy. Now if you, the different ways these As and *s can be arranged, you will notice that each arrangement corresponds to a specific distribution. There are 5 As and 3 *s. The number of ways you can arrange them is [8!/(3!5!)]
Generalised form for n = number of identical objects, and r = number of children is n+r-1Cr-1.

This questions can be solved by a property that's useful for calculating WHOLE NUMBER Solutions (All Non-Negative Solution)

Whole Solution of a linear Equation a+b+c+d+e+.... = n

where we have r variables (a, b, c, d, e)

is (n+r-1)C(r-1)

Here we have, a+b+c+d = 5

so Whole no. solution of the equation = (5+4-1)C(4-1) = 8C3 = 56

Answer: option A

Hi,

I have tried writing an explanation in a lucid way. Hope it helps someone.

Hi camlan1990,

This question comes with a bit of a 'twist.' Since the apples are IDENTICAL, you have to be careful about duplicate 'options'...

For example:
Giving 1 apple to child A and 1 apple to child B

is the SAME as....

Giving 1 apple to child B and 1 apple to child A

Since those duplicate options are NOT supposed to be counted twice, the Combination Formula is necessary (the solutions by icefrog and ankuragarwal1301 showcase that math).

GMAT assassins aren't born, they're made,
Rich

Does this specifically include zero, as in zero apples in a basket?

If so, what formula would you use in the event that everyone had to have at least 1 apple? Can you merely reduce "n" by 1, since n cant be zero? Suppose this there the case with 20 apples and 4 baskets.

I know a simple way to solve this problem.

Consider it as an equation : A + B + C + D = 5. Where letters represent children and 5 = Apples.

Just apply the formula : n+r-1Cr-1

where r = children and N = 5 i.e Apples.
we will get answer as 56.

Remember : This formula is used because question mentions that some children may get 0 apple. If this was not mentioned, there is a different formula to solve the question.

Asked: In how many ways can 5 apples (identical) be distributed among 4 children? (Some children may get no apples.)

AAAAA|||
There are 5 apples and 3 partitions.

Number of ways = 8C3 = 56

IMO A

Can someone direct me to some post or link to understand how to approach such questions. ?

Approach 1:
- memorize the formula ---> (n + k - 1)! / k! * (n - 1)!
- be able to spot exactly when you can apply it (as in this case)
- solve the problem in 45 sec max

Approach 2:
-understand the problem
-solve
-probably 1.30 min gone

here is the second one:

You can divide the 5 IDENTICAL apples in the following ways:

5 - 0 - 0 - 0 4 ways to do that
4 - 1 - 0 - 0 12 ways to do that
3 - 2 - 0 - 0 12 ways to do that
3 - 1 - 1 - 0 12 ways to do that
2 - 2 - 1 - 0 12 ways to do that
2 - 1 - 1 - 1 4 ways to do that

TOTAL 56 ways

why can we use the combinations with repetitions formula?

it looks very different than this question that typically uses combinations with reptition

Example of what i typically consider as C w rep question:
Let us say there are five flavors of icecream: banana, chocolate, lemon, strawberry and vanilla.
We can have three scoops. How many variations will there be?

@veritasprep

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