#Combinatorics

Question

#COMB

There are $4$ boys and $6$ prizes are to be distributed among them such that each has at least $1$ prize. How many ways that can be done?

My solution:
$\text{Case 1 : 3 1 1 1}$

$\text{Case 2 : 2 2 1 11}$

 $\text{Case 1 : C(6,3) * 4! = 480}$

 $\text{Case 2 : C(6,2) * C(4,2) * 4!/2! = 1080}$

 $\text{Case 1 + Case 2 =1560}$
 

My doubt is in the second case, am I not considering the prizes to be indistinguishable? I am confused in this regard. Please help me clear this doubt.

Answer 1

No you are right,this is distinguishable distribution,if case 2 is indistinguishable distribution then you have to  divide (2!×2!) As we can't distinguish between 2,2 and 1,1 so we have to remove those,checc my solution

Comments

Case 1 can be like this

At first give 1 prize to each of them and then give last 2 item to any 2 student 1 -1 each

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It can be done like

$\binom{6}{4}\times 4!\times 6\times 2$

Is it not?

Say, A,B,C,D each got 1 at first

then $\left ( A,B \right ),\left ( A,C \right ),(A,D),\left ( B,C \right ),\left ( B,D \right ),\left ( C,D \right )$

And they can be arrange in 2 ways

Is it not? Anything wrong?

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Sorry but I don't understand (6c4 ×4! ×6×2) means

Once check my solution or abhishek das ,,if u are not able to understand those solutions then we will figure out again.

Answer 2

If prizes are distinguishable-

Case 1-

(1,2,2,1)

4C2 * 6C4 * (4!)/2!2! * 2! = 1080

Case 2-

(1,1,1,3)

4C1 * 6C3 * ((3!)/3!) * 3! = 480

so, 1080+480 = 1560

If prizes are indistinguishable-

Case 1-

(1,2,2,1)

We just need to select 2 out of 4 persons who are getting 2 prizes each,cause selecting indistinguishable prizes won't make any sense as all are same hence there is only one-way,so 4C2 * 1 = 6

case 2-

(3,1,1,1)

Similarly, pick 1 person out of 4 who's getting 3 prizes = 4C3 = 4

Hence, 6+4 = 10

Or,you can apply a formula for indistinguishable-items i.e, (n-1)C(r-1) = 5C3 = 10