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Question

Let Xn = {1,2,3,.n) and let a subset A of Xn be chosen so that every pair of elements of A differ by at least 3. (For example, if n = 5, A can be ϕ, {2} or {1,5} among others). When n = 10, let the probability that 1A be p and let the probability that 2A be q. Then

A
p>q and pq=16
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B
p<q and qp=16
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C
p>q and pq=110
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D
p<q and qp=110
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Solution

The correct option is B p>q and pq=110
When n = 10, Let Ar be no. of ways of selecting r numbers
No. of selection of A is
=n(A0)+n(A1)+n(A2)+n(A3)+n(A4)
1+10+(7+6+5+.+1)+(4+3+2+1)+(3+2+1)+(2+1)+1+1
=11+7.82+10+6+3+1+1=60
N(p) = n (no. of ways 1 is selected) = 1 + 7 + 4 + 3 + 2 + 1 + 1 = 19
N(q) = n (no. of ways 2 is selected) = 1 + 6 + 3 + 2 + 1 = 13
So. p=1960 and q=1360
pq=110

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