# Permutations or Combinations? 476

• Sep 20th 2008, 08:57 AM
fabxx
Permutations or Combinations? 476
Page 476
5 Different colored (black, red, blue, green, and yellow) cards are placed in a row so that the black card is never at either end, how many different arrangements are possible? I don't get why the correct answer is 72.

• Sep 20th 2008, 10:04 AM
Jhevon
Quote:

Originally Posted by fabxx
Page 476
5 Different colored (black, red, blue, green, and yellow) cards are placed in a row so that the black card is never at either end, how many different arrangements are possible? I don't get why the correct answer is 72.

the black card can be in one of 3 positions. 2nd, 3rd, or 4th. for each of those positions, the other 4 cards can be permuted among the other four positions.

hence, the number of arrangements is 3*4! = 72
• Sep 20th 2008, 10:06 AM
Kai
Hello Fabxx,

I believe there are other ways to do it, but i did like this...

First calculate total number of arrangements possible : 5!=120

Then calculate the number of arrangements for black to be first: 1*4!=24

And now the number of arrangements for black to be last : 4!*1=24

So Ur answer would be 120-(24+24)= 72