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# any one good at permutations or combinations?

how do i solvefor these?

4!/ 6! ?

(n+2)!/ n!

(2n-1)!/ (2n+1)!

(2n+2)!/ (2n)!

### 4 Answers

- 1 decade agoFavorite Answer
Expanding upon the previous person's answers,

For the first one, you know that

4! = 4 * 3 * 2 * 1

6! = 6 * 5 * 4 * 3 * 2 * 1

If you divide them, everything cancels out except for 5 and 6 on the bottom, so it's 1 / 30.

You can also see that 6! is just 6 * 5 * 4!, which will help with the later problems.

(n + 2)! = (n+2) * (n+1) * n * (n-1) * .... * 1

n! = n * (n-1) * ... * 1

(n + 2)! = (n + 2) * (n + 1) * n!, so it's

(n + 2) * (n + 1).

You can figure the rest out from there.

- 1 decade ago
4!/6! = 4!/(4!*5*6)= 1/(5*6) = 1/30

(n+2)!/n!= [(n!)*(n+1)*(n+2)]/(n!) = (n+1)(n+2)

(2n-1)!/(2n+1)!= (2n-1)!/[(2n-1)!*(2n)(2n+1)=1/[(2n)(2n+1)]

(2n+2)!/(2n)! = [(2n)!(2n+1)(2n+2)]/(2n)! = (2n+1)(2n+2)

- pbb1001Lv 51 decade ago
These are straight-out calculations!

I will do the first two.

4!/ 6! = 4!/(6*5*4!), right? Cancel 4! or

= 1/(5*6)

= 1/30

(n+2)!/ n! , same thing, write as

((n+2)(n+1)n!)/n!, cancel n! or

= (n+2)(n+1).

Do the rest.

- sahsjingLv 71 decade ago
4!/ 6! = 1/30

(n+2)!/ n! = (n+2)(n+1)

(2n-1)!/ (2n+1)! = 1/[(2n+1)(2n)]

(2n+2)!/ (2n)! = (2n+2)(2n+1)