A teacher will divide her class of 17 students into four groups to work on projects. One group will have 5 students and the other three groups will have 4 students. How many ways can students be assigned to group if:
All group work on the same project?
since there's no order i think the right answer would be :
$$ \frac{\binom{17}{5 ,4,4,4}}{4!}$$
But in my notes the professor said $$ \frac{\binom{17}{5 ,4,4,4}}{3!}$$
I don't see any particular reason.
Maybe it is because we're trying to make the 3 groups of four look the same ?
Is a multinomial coefficients used to distribute those 17 people with taking in account that the 4 groups are different, and dividing that by 3! means we can arrange those groups in 3! ways ?
Best Answer
Think of it this way. Arrange the students in a line (there are $17!$ ways to create this permutation):
$$ \begin{array}{ccccccccccccccccc} s_1 & s_2 & s_3 & s_4 & s_5 & s_6 & s_7 & s_8 & s_9 & s_{10} & s_{11} & s_{12} & s_{13} & s_{14} & s_{15} & s_{16} & s_{17} \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \end{array} $$
Then create the groups by taking the first 4, second 4, third 4 and last 5:
$$ \begin{array}{cccc|cccc|cccc|ccccc} s_1 & s_2 & s_3 & s_4 & s_5 & s_6 & s_7 & s_8 & s_9 & s_{10} & s_{11} & s_{12} & s_{13} & s_{14} & s_{15} & s_{16} & s_{17} \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \end{array} $$
Every way to group the students can be created this way. The key question is: how many permutations lead to the same set of groups?
We see of course that we can permute the people in any of the 4 groups and it doesn't change the group. Example:
$$ \begin{array}{cccc|cccc|cccc|ccccc} s_3 & s_2 & s_4 & s_1 & s_6 & s_5 & s_7 & s_8 & s_9 & s_{10} & s_{12} & s_{11} & s_{14} & s_{13} & s_{17} & s_{15} & s_{16} \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \end{array} $$
There are $4!$ ways to permute each of the first three groups and $5!$ ways to permute the fourth group. This brings us down to $\frac{17!}{4!4!4!5!}$.
Also note, that when we do this, there is an inherit order to the three groups of $4$. Specifically, things like this:
$$ \begin{array}{cccc|cccc|cccc|ccccc} s_5 & s_6 & s_7 & s_8 & s_1 & s_2 & s_3 & s_4 & s_9 & s_{10} & s_{11} & s_{12} & s_{13} & s_{14} & s_{15} & s_{16} & s_{17} \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \end{array} $$
is considered different. There are $3!$ ways to permute the groups of $4$ leading to the final answer of
$$ \frac{17!}{4!4!4!5!}/3!. $$
Note that we cannot swap a group of $4$ with a group of $5$ because that would require moving the bars. The goal we started with is to count how many of the $17!$ permutations lead to the same groups of students by placing the bars after the fourth, eighth and twelfth student.