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How many options are there for 15 student to divide into 3 equal sized groups?

Now i know the soultion is $\;\dfrac {15!}{5!5!5!3!}\;$ but i can't understand why.

Can anyone please enlighten me?

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Consider the following points:

  1. Suppose that we line up the students in a line and choose the first 5 students to belong to group 1, the next 5 to group 2 and the last 5 to group 3.

  2. There are 15! ways to line up students in a line.

  3. However, in any particular line up, the order of the students in each group does not matter. Thus, for each line up, we have 5! 5! 5! ways of arranging the students in each group. Therefore, the number of ways to arrange students in 3 equal groups is: $$\frac{15!}{5! 5! 5!}$$

  4. But, the labeling of the groups also does not matter. We can label the group 1 to be group 2 and group 2 to be group 3 and so on. We have 3! ways to label the groups. Thus, the final number of ways to arrange students in 3 equal groups is:

$$\frac{15!}{5! 5! 5! 3!}$$

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  • $\begingroup$ Assuming you want the case in #3 where the order of the groups matters, does the sum of values in the denominator need to equal the size of the total group? $\endgroup$ – Sebastian Apr 1 at 16:12
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  • There are $\,\binom{15}{5}\,$ ways to choose five students for group ONE,
  • Now, there are $\binom{10}{5}$ ways to choose five students for group TWO,
  • That leaves us with only five students: i.e., hence there is only $\,\binom 55 = 1\,$ way to choose the students for group THREE.

Multiplying these factors: $$\binom{15}{5} \times \binom{10}{5} \times \binom 55= \frac{15!}{10!\,5!}\times \frac{10!}{5!\,5!}\times 1 = \frac{15!}{5!\,5!\,5!}\tag{1}$$

Now we need to divide that total by $\,3!,\,$ since the labeling/ordering of the groups (group ONE, group TWO, group THREE) doesn't matter: since there are $3!$ ways of labeling the groups, we need divide the total given in $(1)$ by $\,3!$.

Hence, our final answer is:

$$\text{There are}\;\left(\frac{15!}{5!\,5!\,5!\,3!}\right)\;\text{ways to divide 15 students into three groups of equal size.}$$

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  • 1
    $\begingroup$ Nice explanation +1 $\endgroup$ – Amzoti May 17 '13 at 0:54
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I actually found a nice way of solving it (After reading the post on the comment):

Say i choose a person first in the line-up, Now that person have $\binom {14}4$ ways of choosing 4 other groups members, Then the next person in the line up has $\binom 94$ ways of choosing other groups members. Finally, the last member has $\binom 44=1$ ways of choosing his group members.

If we multiply all of the above: $\binom {14}4\times$$\binom 94$ and we get the exact same result as stated. (Just written a bit differently.)

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