Came across another textbook question I'm struggling with...

A father buys nine different toys for his four children. In how many ways can he give one child three toys and the remaining three children two toys each?

I have tried the following $4! \cdot \binom{9}{2} \cdot \binom{7}{2} \cdot\binom{5}{2}$,

as in the ways to choose the 4 kids and then to give each one 2 toys with the last kid having 3 toys given to them.

As they are stated to be different I tried permutations as well, but evidently am missing some of the logic.

  • $\begingroup$ Your only error is that you multiplied by $4!$, rather than just $4$. The order of the remaining three kids does not matter. $\endgroup$ – Brian Tung Dec 11 '15 at 0:40

The lucky child can be chosen in $\binom{4}{1}$ ways. Her $3$ toys can be chosen in $\binom{9}{3}$ ways.

That leaves $6$ toys to be distributed among $3$ kids, $2$ to each.

Line up these three kids in order of age. The toys for the oldest can be chosen in $\binom{6}{2}$ ways, and for each way the toys for the next oldest can be chosen in $\binom{4}{2}$ ways, for a total of $\binom{4}{1}\binom{9}{3}\binom{6}{2}\binom{4}{2}$.

There are many other ways to do the counting, which in general will produce answers that may look different than the above answer, but will be numerically equivalent.

Remark: We can choose the lucky child and put her at the end of the line. This can be done in $\binom{4}{1}$ ways. Line up the rest of the children in order of student number. They can be given their toys in $\binom{9}{2}\binom{7}{2}\binom{5}{2}$ ways. That I think was basically your analysis. Almost all your components were right.

  • $\begingroup$ There was another answer that gave the same answer. What happened to it? Missing dollar sign, by the way. $\endgroup$ – Brian Tung Dec 11 '15 at 0:42
  • $\begingroup$ My book says 7560, so I'm assuming this is a typo as this seems to make sense! Thanks for the help :) $\endgroup$ – princesspink Dec 11 '15 at 0:49
  • $\begingroup$ Actually sorry, just the last part is confusing me as it should be $\binom{4}{2}$ then I get 30240? Edit: 30240 is also what I get when I do it the way I originally did with the 4! changed to 4... I'm thinking the textbooks wrong? $\endgroup$ – princesspink Dec 11 '15 at 0:52
  • $\begingroup$ Thanks. I had the right numbers on the line above, but typed incorrectly on the next line. $\endgroup$ – André Nicolas Dec 11 '15 at 0:55

This is a case of overcounting: say the toys are $\{a,b,c,d,e,f,g,h,i\}$ and you have children $\{1,2,3,4\}$. A possible arrangement this counts is $(1,2,3,4)$, and $(\{a,b\},\{c,d\},\{e,f\},\{g,h,i\})$, meaning that the first child gets toys $a$ and $b$, second child gets $c$ and $d$ and so on. However, you could also have the arrangement $(2,1,3,4)$ and $(\{c,d\},\{a,b\},\{e,f\},\{g,h,i\})$ which gives the exact same result.

The problem is that you're accounting for the children moving in different places and the toys moving in different places, but by moving both, you end up just ordering the children when you're just aligning them with the toys. As you see above, the other answers only fix the golden child who gets $3$ toys, as by ordering the partitions of the other $6$ toys, you also decide which child they go to and cover all combinations.


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