Question

How many ways are there to divide $12$ people into a group of $3$, a group of $4$, and a group of $5$, if Henry has to be in the group of $4$

Answer 1

The total number of such groupings of 12 people into groups of 3, 4, and 5, regardless of Henry's placement, is

$\dbinom{12}3 \dbinom94 \dbinom55 = 27,720$

where

$\dbinom nk = \dfrac{n!}{k!(n-k)!}$

is the binomial coefficient. This is because we initially pick 3 people from the total 12, leaving 9 available; then 4 people from these 9, leaving 5 more; then the remaining 5 are all selected.

If Henry claims a spot in the group of 4, then we effectively can only pick 3 people from the 8 non-Henrys to get

$\dbinom{12}3 \dbinom83 \dbinom55 = \boxed{12,320}$

such groupings.