In an election for the managing committee of a reputed club , the no. of candidates contesting election exceeds the no. of members to be elected by r [r is greator than 0] .if a voter can vote in 967 different ways to elect the managing committee by voting atleast 1 of them and can vote in 55 different ways to elect [r-1] candidates by voting in the same manner find the no.of candidates contesting the elections and the no. of candidates losing the election ? give me proper explanation of its answer

Hello student,

Let there be S seats and C candidates, with C - S = r.
For S seats, a voter can vote for
C choose 1, C choose 2, C choose 3, ... etc up to C choose S.
and the sum of those = 967
For r-1 seats, it is similar
but just up to C choose (r-1) and the smaller sum = 55.
Examining Pascal's triangle,
we see that in the 10th row,
10 choose 1 = 10
10 choose 2 = 45
which gives the required sum = 55.
That gives r - 1 = 2 and hence r = 3.
That makes sense since 2^10 =1024
which is the power of two next above 967,
and it is the number of ways to vote for up to 10 candidates
(including voting for none).
So C = 10, r = 3, and S = 7.
We have 10 candidates vying for 7 seats. ← ANSWER
10 choose 1 = 10
10 choose 2 = 45 (sum = 55)
10 choose 3 = 120 (sum = 175)
10 choose 4 = 210 (385)
10 choose 5 = 252 (637)
10 choose 6 = 210 (847)
10 choose 7 = 120 (967)as required.

Thanks and Regards

Shaik Aasif

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