Permutations and combinations class 11

Question

If nC8=nC6, Find nC2.

 

 

Please help me out with proper explanatory steps as in the previous answers...

 

Thanking you,

Diana

Answer 1

Hi Diana,

 

First, it's important to understand what's being asked here. nCr is the abbreviated way of saying "Out of n items, choose r of them (r<n). The order does not matter." So for example, let's say we have the set of numbers {1,2,3}, and I tell you to pick two of them, and their order does not matter- I would write that request as 3C2- and we could write {1,2}, {1,3} and {2,3}. We wouldn't write {2,1}{3,1} or {3,2}since they would be repeats of the same two numbers- this is because again, with combinations, the order does not matter. 

In most other situations, you won't have such easy numbers to deal with. So, let's talk about how we would represent nCr algebraically. Given that we can only choose each item from our sets once (no repeats), we can represent the amount of ways to pick r out of n things with this formula:

nCr=    n!

        ————
         r!(n-r)!

• The n! in the numerator represents all the possibilities of choosing all n items.
• The (n-r)! represents the amount our of n that we would not choose (so if it's 5C3, we're not picking 2 to complete the 5 slots out of 5 available- leaving us with 5*4*3 possible choices). Thus why we would divide it out.
• The r! removes any repeats (since order does not matter in combinations).

With that being said, what we can do now is some algebra to figure out the n we're missing in our original problem:


    n!                n!
———     =    ———
8!(n-8)!        6!(n-6)!


The n! will cancel out, and I'll multiply the denominators on either side to bring them to the top:

6!(n-6)!= 8!(n-8)!

Mind you, when you see something like (n-k)!, you would write it out as (n-k)*(n-k-1)*(n-k-2)*....3*2*1 - as you would have n decrease as much as possible down to 1. We don't know how long that will be, but we would write it symbolically this way. So, I'm going divide on both side by similar terms to get this:

(n-6)!        8!
−−−    = −−−
(n-8)!        6!

I'll stretch out these factorials so you can see what we're now dealing with:

(n-6)*(n-7)*(n-8)*(n-9)*...3*2*1               8*7*6*5*4*3*2*1
−−−−−−−−−−−−−−−−−−−−−  =  −−−−−−−−−−−−−−−
(n-8)*(n-9)*....3*2*1                               6*5*4*3*2*1

We can see some terms are able to cancel out, leaving us with

(n-6)*(n-7) =8*7

(n-6)*(n-7)= 56

We can now do some algebra to find our missing n

n² - 13n +42= 56

n² - 13n - 14=0

From algebra, we can figure out from here that this quadratic equation can reduce to (n-14)(n+1) (-14+1 =-13, 14*-1= -14) we can't have -1 things to choose from, so our missing n is 14.

We can now plug 14 as n for nC2 to get our final result:

 14!                   14!                 14*13
−−−−−−   =  −−−        =   −−−−−−    = 91
2! (14-2)!        2! 12!                  2


Hope this helps!