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

Options

A. `pgtq and p-q=(1)/(6)`
B. `pgtq and q-p=(1)/(6)`
C. `pgtq and p-q=(1)/(10)`
D. `pgtq and q-p=(1)/(10)`

Answer

C

Explanation

where n=10
Let `A_(r)` be no. of ways of selecting r numbers.
No. of selection of A is
`=n(A_(0))+n(A_(1))+n(A_(2))+n(A_(4))`
`=1+10+(7+6+5+ . . .+1)+(4+3+2+1)+(3+2+1)+(2+1)+1`
`=11+(7.8)/(2)+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=(19)/(60)" "q=(13)/(60)`
`p-q=(1)/(10)`