Arranging objects with restrictions

• Oct 5th 2009, 06:37 PM
eeyore
Arranging objects with restrictions
There are 6 men and 9 women and they need to be seated in a row of 15 chairs. How many ways are there to arrange them such that the 6 men sit next to each other?

This is what I have done so far. There are 15! ways of arranging the entire group. Arranging the 6 men is 6!. In addition, there is the question of where you place the group of 6 men. It could be

6 men and then 9 women
a woman, 6 men and then 8 women
2 women, 6 men and then 7 women
...
etc

I do not know how to count these possibilities. I know what the answer is (from answer key) but I just can't arrive at it.

Thanks for taking the time to read my question!
• Oct 5th 2009, 06:50 PM
awkward
Quote:

Originally Posted by eeyore
There are 6 men and 9 women and they need to be seated in a row of 15 chairs. How many ways are there to arrange them such that the 6 men sit next to each other?

This is what I have done so far. There are 15! ways of arranging the entire group. Arranging the 6 men is 6!. In addition, there is the question of where you place the group of 6 men. It could be

6 men and then 9 women
a woman, 6 men and then 8 women
2 women, 6 men and then 7 women
...
etc

I do not know how to count these possibilities. I know what the answer is (from answer key) but I just can't arrive at it.

Thanks for taking the time to read my question!

If we view the chairs as being numbered 1 to 15 from left to right, then the leftmost man of the group of 6 could be in chair 1, 2, 3, ..., 10. So there are 10 ways to place the group of men without regard to the order of the men within the group. Once we have figured out where to place the group of men, there are 6! ways to arrange the men and 9! ways to arrange the women.

So there are $10 \times 6! \times 9!$ ways in all.
• Oct 5th 2009, 07:25 PM
eeyore
That's what I thought at first but the answer says it's 6! 10! / 15!. Are you sure the way the women are arranged shouldn't be considered as well? If you decide to place the men in the first 6 chairs, then the 9 women can be arranged in 9! ways and each of those arrangements is consistent with the requirement that the men are next to each other.
• Oct 5th 2009, 07:28 PM
Soroban
Hello, eeyore!

Quote:

There are 6 men and 9 women to be seated in a row of 15 chairs.
How many ways are there to arrange them such that the 6 men sit next to each other?

Duct-tape the six men together.
. . There are $6!$ possible orders.

Then we have: . $\boxed{MMMMMM}, W, W, W, W, W, W, W, W, W$

With 10 "people" to arrange, there are $10!$ possible orders.

Therefore, there are: . $(6!)(10!) \:=\:2,\!612,\!736,\!000$ ways.

• Oct 5th 2009, 07:44 PM
eeyore
Very neat way of thinking of it. Thank you!