Number of ways in which four different toys and five indistinguishable
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17 Feb 2021, 16:37
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)
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