Find the number of ways n different games can be divided

[chaoseverlasting]

Homework Statement

Find the number of ways n different games can be divided between n different children so that every time, exactly one child gets no game.

Homework Equations

The Attempt at a Solution

Here, since exactly one child gets no games, n games are distributed among (n-1) children which gives one child 2 games. Now, since a child can get 2 games out of n in ^n C_2 and the children can be arranged in n! ways, to total ways to distribute the games is ^n C_2 * n!

I was wondering if the following solution is also correct:
x_1 + x_2 +x_3 +... x_n =n such that exactly one x_i =0 and exactly one x_j =2 where i and j are not the same elements and all other elements are equal to 1.

Therefore, the solution should be:

Coeff. of x^n in ( (x^0)(x^1+x^2)(x^(n-2) )^n

Thats coeff. of x^n in ( (x^0)(x^1+x^2)(x^(n-2) )^n

Is this correct?

 

[AlephZero]

As an independent check, here's a different argument.

Pick one child to receive no games. n choices.

Pick the child who receives 2 games. (n-1) choices.

Pick the two games to be received by that child. n(n-1) choices.

Give each remaining child one game chosen at random. (n-2)! choices.

The above are independent, so the number of combinations = n^2(n-1)^2(n-2)! = n(n-1)n!

 

[chaoseverlasting]

btw, how would you find the coeff. of x^n in that equation though? And thank you Azero, that's the actual given solution.

 

[arunbg]

Pick the two games to be received by that child. n(n-1) choices.

That would be n(n-1)/2 .

Chaoseverlasting, I can't quite see how you came up with the
x_1 + x_2 +x_3 +... x_n =n to solve the problem, since that approach would only give you the number of combinations if the games were not distinct. In this case however, even the games are distinct and thereby increases the no. of combinations.

Cheers

 

[AlephZero]

That would be n(n-1)/2 .

Oops - you are right, of course.