In an election for the managing committee of a reputed club, the numbe

Question

In an election for the managing committee of a reputed club, the number of candidates contesting elections exceeds the number of members to be elected by r (r > 0). If a voter can vote in 967 different ways to elect managing committee by voting at least 1 of them & can vote in 55 different ways to elect (r – 1) candidates by voting in the same manner. What is the number of candidates contesting the election & the number of candidates losing the elections?

A. 8and 3
B. 10 and 3
C. 12 and 3
D. 10 and 4
E. 10 and 5

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A. 8and 3
B. 10 and 3
C. 12 and 3
D. 10 and 4
E. 10 and 5

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chetan2u · Answer

Let us simplify the question by breaking it into variables.

Let the total number of the candidates be n.
The number of members to be elected =n-r
Thus a voter can vote for 1 of them or 2 of them till n-r in 967 ways. 
That is nC1+nC2+nC3+....nC(n-r)=967 ways
 The above sum should immediately get you to the formula 
nC0+nC1+....nCn=2^n 
So we have to get above formula into this form. 
Add nC0 to both sides. 
nC0+ nC1+nC2+nC3+....nC(n-r)=967+nC0=967+1=968....(I)

Now, let us take the second part. 
A voter can vote in 55 different ways to elect (r – 1) candidates by voting in the same manner
So nC1+nC2+...nC(r-1)=55
nC0+nC1+nC2+.....nC(r-1)=55+nC0=55+1=56.....(II)
 The other thing one should know is that nC0=nCn and nC1=nC(n-1) 
Replace these values in (II)
nCn+nC(n-1)+....nC(n-(r-1))=56
Add equation (I) to above equation 
 nC0+nC1+....nC(n-r)+nC(n-(r-1))+.....nC(n-(r-r))=968+56=1024=2^n.......n=10

So total candidates are 10.
10C1+10C2+...10C(r-1)=55
10+45+..=55
So, r-1=2.......r=3

Answer : 10 and 3

B

hr1212 · Answer

Question language can be a little confusing at first glance.

Required knowledge -

nC0 + nC1 + nC2 + nC3 + .... + nCn = 2^n

Let the number of candidates be n and elected candidates be e, then n - e = r

I tried it with intuition and taking answer choices as reference.

967 => This number is closest to 1024, which is  2^{10}

We have 3 answer choices (B, D, E) with n = 10, so with probability in our favor it's best that we assume this to be true and try calculating e or r.

10C0 + 10C1 + nC2 + nC3 + .... + 10C10 = 2^{10} 
We now need to exclude 10C0, and few other combinations from the end to equate to 967

10C1 + 10C2 + 10C3 + .... + 10Ce = 1024 - 10C0 - 10C10 - 10C9 - .... - 10C(e+1)  --- (1)
As we can see r is a relatively small number (3, 4, 5) from the option choice, we just need to subtract limited combinations

1024 - 10C0 - 10C10 - 10C9 - 10C8 --- (2)
1024 - 1 - 1 - 10 - 45
967 => Required total combinations

From 1 and 2, e + 1 = 8 => e = 7

r = n - e = 10 - 7 = 3

There's high probability that this is the correct answer choice. If time permits and want to verify the same from the additional condition - 55 different ways to elect (r – 1) candidates

which means, 10C1 + 10C2 = 55?
This can be easily verified and we can confirm that we have selected the right choice.

Answer: B

In an election for the managing committee of a reputed club, the numbe

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