gig92 wrote:In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?

OA-Later

This is analogous to arranging six letters with repetitions: GGGRRR

We can always take this = 6!/ (3! × 3!)

= [spoiler]20[/spoiler]

sanju09 wrote:

gig92 wrote:In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?

OA-Later

This is analogous to arranging six letters with repetitions: GGGRRR

We can always take this = 6!/ (3! × 3!)

= [spoiler]20[/spoiler]

Thanks, I think I need some good sleep!! Morning math was not a good idea!!

Why does the color of the shirts matter? I worked this out to be 6! since there is nothing indicating that the colors of the shirts make a difference. In other words, how many ways can 6 kids get 6 shirts. What am I missing?

according to if the color of the shirts doesn't matter then 6C6 or simply 6 shirts and not 6!
6! would be if you have one shirt for six kids and distribute this 6! ways or you've got 6 shirts and only one kid you could distribute this 6! by making different in each and every shirt and each and every kid receiving a shirt

here the questions puts down with black on white --> two colors (feel the difference?) three of each color (?) distributed to all kids (?)

first distribute 3 shirts of the same color (say green) and the remaining "undressed" kids should receive other color shirt, hence 6C3=6!/(3!*(6-3)!)=20. Our distribution of one color shirts (three to each ~3~ kids) leaves other three kids expecting different color shirts. In total 20 ways make up our distribution.

jk2010 wrote:Why does the color of the shirts matter? I worked this out to be 6! since there is nothing indicating that the colors of the shirts make a difference. In other words, how many ways can 6 kids get 6 shirts. What am I missing?

Thanks... that makes sense. I guess they include the colors for a reason!