I’ll give it a try....

I believe the question is saying “at least one toy and at least one marble”


(1st) distributing the 4 different toys to 3 different people

Each person must receive at least 1 of the 4 toys. This means one person will receive 2 toys and the other two people will receive just 1 toy.


[2 ; 1 ; 1]

Step 1: distribute the different toys into identical stacks as listed above

“4 choose 2” * “2 choose 1” * 1 choose 1” = 4! / (2! * 1! * 1!)

However, since there are identical amounts of toys placed in the stacks that contain 1, we will have repeat counts. To remove the overcounting, we need to divide by 2!

End up with: 4! / (2! * 2!*) = 6 ways

AND

Step 2: Because we have 3 distinct groups of people, we need to arrange the “stacks” among the 3 people. Because each stack has different items, it matters which stack with 1 toy goes to which person.

We can arrange the 3 different stacks among 3 different people in 3! Ways

Toys can be distributed in: 6 * 3! = 36 ways

And

(2nd) distribute 5 identical marbles among 3 distinct “groups” (people)

Letting the 3 different people be variables A B and C

A + B + C = 5

If each person must receive 1, then we are left with the linear equation:

a + b + c = 2

No of ways to distribute the identical marbles = 4! / (2! * 2!) = 6 ways


(36 ways to distribute the 4 toys) * (6 ways to distribute the 5 identical marbles) = 216 ways

D 216

Hopefully I didn’t miss anything (and read the question correctly)

Posted from my mobile device

The 4 toys can be distributed in 2 , 1, 1 ways where each gets at least 1 toy.

The number of ways of doing this is 4C2 * 2C1 * 1C1 = (4 * 3/2) * 2 * 1 = 12 ways

Now the number of ways where 2,1,1 can be rearranged = 3!/2! = 3 ways (211) (121)(112)

Total ways of distributing the toys = 12 * 3 = 36 ways


For the marbles, since they are the same, let us give away 1 marble to each. We are left with 2 marbles. The number of ways of distributing them is

2, 0, 0 = 3!/2! = 3 ways and

1, 1, 0 = 3!/2! = 3 ways

The total ways = 3 + 3 = 6


Therefore the total distribution can be done in 36 * 6 = 216 ways


Option D

Arun Kumar
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Solution:

Let’s begin with the number of ways to distribute the toys. Since there are three children and four toys, and since each child will receive at least one toy, then one child must get two toys and the remaining two children must get one toy. If Amar is the child to get two toys, there are 4C2 = 6 ways to choose the toys that he will receive. Akbar will get one of the two remaining toys, so there are 2 ways to choose the toy that he will receive. Anthony will receive the remaining toy. Thus, there are 6 x 2 = 12 ways in which Amar gets two toys and the remaining children get one toy each.

The number of ways where Amar gets two toys and the number of ways where Anthony gets two toys are also 12; both can be calculated with a similar reasoning as above. Thus, there are 12 + 12 + 12 = 36 ways to distribute the toys.

Now, let’s calculate the number of ways to distribute the marbles. Since the marbles are identical and since each child will receive at least one marble, let’s give each child one marble and determine the number of ways we can distribute the remaining two marbles among the three children such that each child gets 0, 1 or 2 marbles. Notice that there are two scenarios: 1) one child gets two marbles and the remaining children don’t get any marbles, or 2) two children get one marble each and the remaining child doesn’t get any marbles. In the former case, there are three choices for the child to get two marbles, and in the latter case, there are three choices for the child to get zero marbles. Thus, there are 3 + 3 = 6 ways to distribute the two marbles among the children after each child gets one marble.

Since there are 36 ways to distribute the toys and since there are 6 ways to distribute the marbles, there are 36 x 6 = 216 ways to distribute the toys and the marbles.

Answer: D

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