Combinations and Permutation Problems!

Question

In how many ways can the letters of the word (number) be arranged if the e and r cannot be side by side?

 

In how many arrangements can 6 women and 6 men be placed in a line if all the women line up first?

 

In how many ways can you select a committee of 3 Democrats and 3 Republicans from a group containing 24 Democrats and 13 Republicans?

 

In how many ways can you select a group of 7 red cards and 4 black cards from a deck containing 10 red cards and 9 black cards?

Answer 1

Hi Shelby

 

For the first one: first find the total number of ways the letters can be arranged. This is just 6! = 720. Then think of "er" as one letter, then "re" as one letter, and find the ways to arrange for each. If you think of "er" as one letter, it's like you're arranging five letters, so you can do that in 5! = 120 different ways; and same for "re": 5! = 120. Therefore the answer is 720 - (2*120) = 480.

 

For the second question, you just have two blocks: one of women, one of men. The number of ways to arrange the men is 6! = 720, and same for the women: 6! = 720. Then you put the two blocks next to each other: for each arrangement of men you have 720 arrangements of women. So the total possibilities is 720*720 = 518,400.

 

Third question - now we're doing combinations, not permutations. You pick a group of 3 Democrats from 24. This is 24C3 or (24!)/(3!21!) = 2,024. Picking 3 Republicans from 13: 13C3 or (13!)/(3!10!) = 286. For each of the 2,024 groups of Democrats you have 286 possible Republican groups, so total is 2,024 * 286 possibilities, or 578,864.

 

Fourth question is basically the same. For the red cards it's 10C7 or (10!)/(7!3!) = 120. For the black cards it's 9C4 or (9!)/(4!5!) = 126. Altogether you have 120 * 126 = 15,120 possibilities.