The set S has 36 different subsets each of which contains

Question

The set S has 36 different subsets each of which contains exactly two elements. How many subsets of S could contain exactly three elements each?

A. 24
B. 42
C. 54
D. 72
E. 84

BrushMyQuant · Accepted Answer

nC2 = 36
=> n*(n-1)/2 = 36

=> n = 9

nC3 = 9C3 = 84

So, Answer is E.

Hope it helps!

arindamsur · Answer

nC2 = 36
=> n*(n-1)/2 = 36

Explain this portion please.

BrushMyQuant · Answer

n*(n-1) = 36*2 = 72
There are two ways of solving this
Method1:
n*(n-1)=72 = 9*8
n*(n-1) = 9*(9-1)
=> n =9

or n*(n-1) = -9 * -8
=> n*(n-1) = (-8)*(-8-1)
=> n = -8
n cant be negative so, n = 9

Method 2:
n*(n-1) = 72
=> n^2 - n - 72 =0
=> n^2 -9n + 8n - 72 = 0
=> (n-9)*(n+8) = 0
=> n= -8, 9
n cant be negative, so n =9

hope it helps!

arindamsur · Answer

I dont understand how do you derived n*(n-1)/2 from nC2 
=>

paandey · Answer

Why nc2 = 36 ?

What I understand from question is that s = ((a1,b1),(a2,b2),(a3,b3),(a4,b4)...............(a36,b36))
 
Is this wrong understanding of question ? Kindly help

BrushMyQuant · Answer

The Set S has "n" elements {x1,x1,...,xn}

Out of these "n" elements if we try to make subset of two element each then we can make 36 such subsets, which are correctly denoted by you as (a1,b1),(a2,b2),(a3,b3),(a4,b4)...............(a36,b36)
So, S' = (a1,b1),(a2,b2),(a3,b3),(a4,b4)...............(a36,b36)
(Note a ' after S in previous line)

Now, to find out the number of elements in the set S we are given by we can make 36 subsets of two elements each,
So number of subsets of two element each which can be formed from a Set S which has n elements = nC2

read the second half of the question which says "How many subsets of S could contain exactly three elements each"
So, S is the set of "n" elements and not the Set of subsets of two elements

Hope it helps!

Temurkhon · Answer

It was problematic for me to realize that n!/(n-2)!=n(n-1)

KarishmaB · Answer

The question tells us that when we select exactly 2 elements from a set of N elements, we can do it in 36 different ways (hence 36 subsets with 2 elements)

This means NC2 = 36

N*(N - 1)/2 = 36 (using combinatorics)

N*(N - 1) = 72 = 9*8

Hence, N = 9

Now, we need to find in how many ways we can make a subset of exactly 3 elements from N elements.

We can do it in 9C3 ways = (9*8*7) / (3*2*1) = 84 ways

Answer (E)

exc4libur · Answer

n/(n-2)!2!=36, n(n-1)(n-2)!/(n-2)!=72
n(n-1)=72, consecutive int, 9(8)=72
n=9

9!/3!6!=9*8*7/6=12*7=84

Ans (E)

BrushMyQuant · Answer

Just Elaborating on nC2 part of the problem:

nC2 =  \frac{n!}{ (n-2)! * 2! } 
 =  \frac{n*(n-1)*(n-2)*(n-3)*.....*1 }{ (n-2)! * 2!} 
 =  \frac{n*(n-1)*(n-2)! }{ (n-2)! * 2!}    
(n-2)! will cancel out from numerator and denominator
 =  \frac{n*(n-1)}{2 }

Hope it helps!

SmileAndSolve · Answer

How do we know that the two selected numbers are distinct?

Bunuel · Answer

A set, by definition, is a collection of distinct items, so any subset of a set will also consist of distinct items.

bumpbot · Answer

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The set S has 36 different subsets each of which contains

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