# Ways of arranging people in a photo

• Sep 15th 2010, 11:35 AM
mngeow
Ways of arranging people in a photo
7) I have a group of 10 people and I need to select 6 of them to pose for a photograph.Two of the 10 people are named Ken and Karen.

(a)How many ways can I arrange 6 people out of the group of 10 for the photograph if ken and karen must be in the photograph?

(b)How many ways can I arrange 6 people out of the group of 10 for the photograph if Ken and Karen must be in the photograph and they must appear next to each other in the photo?

• Sep 15th 2010, 11:42 AM
undefined
Quote:

Originally Posted by mngeow
7) I have a group of 10 people and I need to select 6 of them to pose for a photograph.Two of the 10 people are named Ken and Karen.

(a)How many ways can I arrange 6 people out of the group of 10 for the photograph if ken and karen must be in the photograph?

(b)How many ways can I arrange 6 people out of the group of 10 for the photograph if Ken and Karen must be in the photograph and they must appear next to each other in the photo?

Have you learned binomial coefficients and permutations?

(a) After the requirement that ken and karen are chosen, you must select 4 people from 8. I believe "arrange" indicates that order matters, so you must multiply the number of combinations by 6!.

(b) It's like (a) except that instead of multiplying by 6! you multiply by 5!2!, try to think why.
• Sep 15th 2010, 12:12 PM
mngeow
Quote:

Originally Posted by undefined
Have you learned binomial coefficients and permutations?

(a) After the requirement that ken and karen are chosen, you must select 4 people from 8. I believe "arrange" indicates that order matters, so you must multiply the number of combinations by 6!.

(b) It's like (a) except that instead of multiplying by 6! you multiply by 5!2!, try to think why.

Yes I have learned permutations and combinations.

Ok I get the part about selecting 4 people from 8,so there's no step to make sure Ken and Karen are selected?We just assume that they are and work from there?I'm also not too sure why you multiply 8C4 by 6!.
• Sep 15th 2010, 12:28 PM
undefined
Quote:

Originally Posted by mngeow
Yes I have learned permutations and combinations.

Ok I get the part about selecting 4 people from 8,so there's no step to make sure Ken and Karen are selected?We just assume that they are and work from there?I'm also not too sure why you multiply 8C4 by 6!.

You can think of it as selecting Ken and Karen first and then choosing the other 4 afterwards. There's only one way to choose Ken and Karen, so it's 1 * 8C4.

But this gives us just the number of ways to select those people. The question statement has "arrange" meaning, I assume, in a line (as in, from left to right) where a different order constitutes a different arrangement. So for each combination, you multiply by 6! to get permutations.
• Sep 15th 2010, 12:45 PM
Plato
Quote:

Originally Posted by mngeow
7) I have a group of 10 people and I need to select 6 of them to pose for a photograph.Two of the 10 people are named Ken and Karen.

(a)How many ways can I arrange 6 people out of the group of 10 for the photograph if ken and karen must be in the photograph?

(b)How many ways can I arrange 6 people out of the group of 10 for the photograph if Ken and Karen must be in the photograph and they must appear next to each other in the photo?

I assume a line of six.

For part a) I would do this: $\displaystyle \displaystyle \binom{8}{4}\cdot 6!$

For part b) I would do this: $\displaystyle \displaystyle \binom{8}{4}\cdot 2 \cdot 5!$

Do you see why?
• Sep 15th 2010, 02:16 PM
Quote:

Originally Posted by mngeow
Yes I have learned permutations and combinations.

Ok I get the part about selecting 4 people from 8,

so there's no step to make sure Ken and Karen are selected? We just assume that they are and work from there?

I'm also not too sure why you multiply 8C4 by 6!.

Ken and Karen have already been selected to be in the photograph.
Hence the problem reduces to selecting 4 people from the remaining 8 to join them in the picture.

Multiplying by 6! then accounts for each person being in any of 6 positions.
That's the number of photos that can be taken with those 6 "particular" people shuffled around in a line.
Multiply by the number of selections since that's the number of "different groups of 4" that can be selected from 8.
• Sep 15th 2010, 08:02 PM
mngeow
I see,thats how it works.I think i get it now,thanks alot guys!
• Aug 5th 2012, 03:50 AM
PTZmm
Re: Ways of arranging people in a photo
You guys rock! Now, I am quite clear of that problem. Thanks!
• Aug 26th 2012, 09:59 PM
louisejane
Re: Ways of arranging people in a photo