Can I solve a Combination Question Algebraically?

Question

For instance, can I find out what n is if nC4 is 5 times nC3? If so how?

Answer 1

use the formula nCr is n!/(n-r)! where n! = 1*2*3*...*n

nC4 = 5*nC3 means n!/(n-4)! = 5*n!/(n-3)!

so (n-4)!*5 = (n-3)!
n-3 = 5
n = 8

Answer 2

Whether you can solve it or not is another issue altogether, but you can at least state the definitions of nC4 algebraically.

Recall that nCk = n!/[k!(n - k)!]

nC4 = n!/[4!(n - 4)!]


If we were to solve your hypothetical equation
nC4 = 5(nC3), then we would be solving this:

n!/[4!(n - 4)!] = 5(n!/[3!(n - 3)!]

We can cross multiply, to obtain

n! (3!) (n - 3)! = 5 n! (4!) (n - 4)!

Note that we can divide both sides by n!, to get

(3!) (n - 3)! = 5 (4!) (n - 4)!

We can also divide both sides by (n - 3)!, to get
(3!) = 5 (4!) [(n - 4)!]/[(n - 3)!]

Note that a factorial can always be stated as the following:

m! = m * (m - 1)!. We're going to do this for (n - 4)!.

3! = 5 (4!) [(n - 4) (n - 3)! / (n - 3)! ]

Now, we can cancel out the (n - 3)!, to get

3! = 5 (4!) (n - 4)

And now we solve for n normally.

3! = 5 (4!) (n - 4)

(3 * 2 * 1) = 5 (4 * 3 * 2 * 1) (n - 4)
6 = 5 (24) (n - 4)
6 = 120 (n - 4)
6/120 = n - 4
1/20 = n - 4
1/20 + 4 = n
81/20 = n

Edit: May be incorrect. But see how it nicely worked out to a single n to deal with.

Edit#2: On second thought, it might actually be the correct answer.

Answer 3

81/20 seems to be correct, except that I would question what it would mean to have a non-integral value for n where you have nCr. Therefore, I would say there is no answer.

Answer 4

actually nCr=n!/(n-r)!*r!
that's the way u can find n
nC4=5(nC3)
n!/(n-4)!*4!=5(n!/(n-3)!*3!)
1/(n-4)(n-3)!4*3!=5(1/(n-3)!*3!)
1/4n-16=5
1=20n-80
20n=81
n=81/20

Answer 5

nCr = nPr/r! , P stands for probability
nCr = n! / r(n-r)!