I have this True/False problem to solve. The number of ways of distributing $8$ identical red balls and $7$ distinct pens in $5$ distinct boxes is greater than the number of ways of arranging $12$ persons into (at most) $4$ groups. (T/F)

I would guess the former number would be $5^{7} {8+4 \choose 4}$ because $8$ balls are identical, $7$ pens are distinct.

However, for the 2nd part I do not know the exact answer. I estimated it by saying that the exact number would be less than that, obtained from dividing the number of groups into cases of 1,2,3,4, and because there are overlaps in the cases, which is something like $4^{12}-3^{12}+2^{12}-1^{12}$, to compromise for the overlaps in the number of groups during consideration.

However, I would think the order of placement in the group matters, hence there might be some kind of duplication factored in the calculation.

Another alternative would be to divide each number of groups further into giving the size of each group, so we are actually counting the number of $4$-element sets that is possible within each number of groups. However, that itself is very tedious as well.

Is there a better way of estimation or even better still a way to exactly divide into cases, because I'm not very good at that.

  • $\begingroup$ The second number is the number of solutions of the equation $x_1+x_2+x_3+x_4=124 with $x_i \geq 0$. $\endgroup$ – baharampuri Nov 23 '18 at 9:42
  • $\begingroup$ @baharampuri Sorry do you mind clarifying the latex? $\endgroup$ – Prashin Jeevaganth Nov 23 '18 at 9:53
  • $\begingroup$ Sorry "with $x_i \geq 0$" $\endgroup$ – baharampuri Nov 24 '18 at 6:53

Your calculation of the number of ways of distributing $8$ identical red balls and $7$ distinct pens to five distinct boxes is correct.

If the groups are labeled, then the number of ways of placing the people in up to four groups is $4^{12}$. If the groups are not labeled, there are fewer ways to place the people in up to four groups.

Observe that $4^{12} < 5^7\binom{12}{4}$.


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