Question

3 women and 5 men are standing in a line. If no two women may be adjacent to each other, how many distinct line-ups are there?

Answer 1

First place 5 men : 5! ways

Answer 2

After that you can place women in any of the places BETWEEN men : 6P3 ways

Answer 3

So total distinct line-ups = 5!*6P3 = 5! * 6!/3! = 14400

Answer 4

Thanks for that I don't really understand the 6P3 part though

Answer 5

Can you please explain when some xPy is used instead of some xCy? :)

Answer 6

xPy is permutation

xCy is combination

Answer 7

with permutation order matters ( `abcd` is not same as `dcba`)
\[nPr = \dfrac{n!}{(n-r)!}\]

with combination order doesn't matter ( `abcd` is same as `dcba`)
\[nCr = \dfrac{n!}{(n-r)!r!}\]

Answer 8

Hmm. But when it comes to application, I often do not know when to use P and C

Answer 9

lets do two quick baby problems

Answer 10

problem1 : Sppose my openstudy password consists of "6 different letters".
You want to break my password for some reason. so you want to know how many different ways that a 6 letter string can be made using 26 alphabets.

question to you - does the order of letters in the string matter here ?