Test: Combinations - Commerce MCQ

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 = ¹⁸C₅

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:
⁶C₁ + ⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆ 
= 6 + 15 + 20 + 15 + 6 + 1 
= 63

ⁿC₄ = 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

⁸C_r = ⁸C_p = ⁸C^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.

For the 1st point chords can be drawn with joining 9 other points.
For the 2nd point distinct chords can be drawn with joining 8 other points.
 For the 9th point distinct chords can be drawn with joining 1 other point. 
So, total no. of points = 9 + 8 + 7 + . . . . . . + 1 = 45 

At least one yellow chair means “total way no yellow chair”
Total ways to select 3 chair form the total (2+3+4) = ⁹C₃

Now, we dont want even one yellow chair.
So we should select 3 chairs from 6 chairs (2 green & 4 blue) = ⁶C₃

Hence, ways to select at least one yellow chair = 84 - 20 = 64 ways.

²⁰C_r = ²⁰C_r+2
ⁿC_r = ⁿC_n-r
=> r + (n−r) = n
⇒ r + r + 2 = 20
⇒ 2r = 18
⇒ r = 9
^rC₆ = ⁹C₆
⇒ 9!/(6!*3!)
= 84

Total number of girls in the group = 4
Total number of boys in the group = 7
Probability of selecting atleast three boy and two girl = ⁴C₂ × ⁷C₃
= 210

8C5 
= 8!/5!*3!
= 8*7*6*5!/5!*3!
= 56

Triangle formed = mC3
Therefore, 4 points on circle, m = 4
4C3 = 4C1
⇒ 4

out of 7 persons , we have to select exactly 3 girls . hence, it's clear that remaining 4 persons will be boys .
 
Now, number of ways taken 4 boys from 9 boys = 9C4
and number of ways taken 3 girls from 4 girls = 4C3
 
hence, by fundamental principle of counting ,
total number of ways for making the committee = 9C4 × 4C3
= 9!/5!.4! × 4!/3!
= 9×8×7×6/3×2×1
= 9×8×7
= 504 ways

For a convex n-sided polygon, there are n vertices, and from each vertex you can draw n-3 diagonals, so the total number of diagonals that can be drawn is n(n-3).The number of diagonals in a polygon = n(n-3)/2, where n is the number of polygon sides.