Permutations and Combinations Test - 6

nC12 = nC8
n/ 12 (n - 12) = n/8 (n - 8)
12 = n - 8
n = 12 + 8 = 20
n = 20.

Out of 20 professors, excluding the principal and vice-principal, 18 professors are left.
No. of ways of selecting 5 professors out of 18 = ¹⁸ C₅

No. of diagonals in a n-sided polygon = n *(n –3)/2     
No. of diagonals in a hexagon = 6 *(6 –3)/2 = 9

She has 6 friends and he wants to invite one or more. That is the same as saying he wants to invite at least 1 of his friends.
 
So, the number of ways he could do this is:
Invite only one friend
Invite any two friends
Invite any three friends
Invite any four friends
Invite any five friends
Invite all six friends
This can be thought of in terms of combinations. Inviting  r  friends out of  n  is same as choosing  r  friends out of  n . So, we can write the possibilities as:
⁶ C₁ +⁶ C₂ + ⁶ C₃ + ⁶ C₄ + ⁶ C₅ + ⁶ C₆  
= 6 + 15 + 20 + 15 + 6 + 1  
= 63

ⁿ C₄ = 495
n!/r!(n-r)! = 495
n!/4!(n-4)! = 495
n!/24(n-4)! = 495
n!/(n-4)! = 495 * 24
n!/(n-4)! = 11850
[n(n-1)(n-2)(n-3)(n-4)!]/(n-4)! = 11850
[n(n-1)(n-2)(n-3)] = 12*11*10*9
n = 12

Given nC8 = nC2
if nCr = nCp
Then either r=p or r=n −p
Thus, nC8 = nC2
8=n −2
10=n
∴n=10

⁸ C_r = ⁸ C_p = ⁸ C^8-p    
So, either r = p or r = 8 –p
i.e. r = p or r + p = 8

18P4/18C4
⇒[18!/14!]/[18!/(4!*14!)]
⇒4!
⇒24

11 players can be selected out of 16 in ¹⁶C₁₁ ways = 16!/(11! 5!) = 4368 ways. Now, If two particular players is to be included and one particular player is to be rejected, then we have to select 9 more players out of 13 in ¹³C₉ ways.
= 13!/(9! 4!) = 715 ways.