# Simple Combinatorics/Problem - Counting Ways to Seat men and women

The Question:

There are 3 men and 3 women to be seated in a row of 10 chairs. In how many different ways can they be seated if one man must be seated at each end of the row.

My attempt:

One man must be seated at each end of the row so really the remaining 1 man and 3 women must be seated in any order in the inner 8 chairs of the 10 chair row.

So to pick the chairs there are $$8C4$$ options, so 70 different ways you can choose the chairs. I think (hopefully) I'm correct in my thinking for this first part but then to do the remaining I'm a bit stuck.

• Generally, when talking about problems involving "men" and "women" each person is different and we can tell them apart based on name, age, hair color, etc... Compare this to a similarly phrased problem where we talk about balls which are generally intended to be indistinguishable apart from color. That all being said, your answer does not yet take into account which of the selected seats is used by each person on an individual level. That is, you only looked at the number of ways of selecting the chairs, not the number of ways of seating the people. – JMoravitz Jun 6 at 12:07

Apply multiplication principle.

• Choose which man sits at the far left

• Choose which man sits at the far right

• Choose which inner seat is used by the remaining man

• Choose which remaining inner seat is used by the oldest woman

• Choose which remaining inner seat is used by the next oldest woman

• Choose which remaining inner seat is used by the final woman

Equivalently it could have been described using the steps:

• Choose which man sits at the far left

• Choose which man sits at the far right

• Choose which four of the remaining eight chairs are used by the remaining people

• Choose the order of the remaining four people for how they sit in the selected chairs

In either phrasing of the solution, you apply multiplication by multiplying the number of options for each step together to arrive at a final answer.

$$3\times 2\times 8\times 7\times 6\times 5 = 3\times 2\times \binom{8}{4}\times 4!$$

As mentioned above in a comment, questions involving ordering and arranging of people we generally assume the people are themselves distinguishable. If the same problem were repeated but talking about balls instead which are ordinarily assumed to be indistinguishable apart from color, then you would just pick which four inner positions were used and then which of those selected positions was the one for the blue ball, not caring explicitly which ball was in which spot.