# Combination and Permutation questions

1. Dec 15, 2004

### yc716

Hi,
Can anyone help me with these permutation/combination questions?

Solve the equation for n:
1. nC4 = 35
2. nC4 = 70

It would be really good if I got the answers with full explanations, a.s.a.p. Thanks.

Last edited: Dec 15, 2004
2. Dec 16, 2004

### matt grime

THere is nothing wrong with trial and error some times.

nCr is increasing (for a fixed r)

Alternatively write nC4=35 out in factorials and get a 4th order polynomial to solve.

You can improve this since it is the same as n(n-1)(n-2)(n-3)=35*24

so n is very close to the 4th root of 35*24

3. Jan 7, 2005

### Dr.ThinkDeep

solve(n*(n-1)*(n-2)*(n-3)=35*24,n); -4, 7, 3/2 + 1/2 I sqrt(111), cc.

solve(n*(n-1)*(n-2)*(n-3)=70*24,n); -5, 8, 3/2 + 1/2 I sqrt(159), cc

I cannot remember the formula for quartic polynomials. Just look up in Abramowich or use Maple.

Last edited: Jan 7, 2005
4. Jan 7, 2005

### HallsofIvy

Staff Emeritus
Since n must be an integer, it's easier to use "trial and error". In particular 6C4= $\frac{(6)(5)}{(2)}$= 15 which is too small while 7C4= $\frac{(7)(6)(5)}{(3)(2)}$= 35. Aha!

8C4= $\frac{(8)(7)(6)(5)}{(4)(3)(2)}$= 70.

5. Jun 29, 2008

### SANCHIT123

yes
nC4=n!/4!.(n-4)!=n(n-1)(n-2)(n-3)[(n-4)!]/(n-4)!.24=35
now n(n-1)(n-2)(n-3)=35*24
n(n-1)(n-2)(n-3)=7*6*5*4
compairing both sides we gwt
n=7
n-1=6 so n=7
ans n=7

6. Jun 29, 2008

### matt grime

This thread is 3.5 years old, doubt that it's going to be of interest to the OP now.

7. Jun 29, 2008

### HallsofIvy

Staff Emeritus
I wondered why I didn't recognize my own response!