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Q.

x1,x2x3,.....,x50are fifty real numbers such that xr<xr+1 for r=1,2,3,...,49. Five numbers out of these are picked up at random. The probability that the five numbers have x20 as the middle number is

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a

 19C2×31C3 50C5

b

 30C2×19C2 50C5

c

 29C2×21C3 50C5

d

 20C2×30C2 50C5

answer is B.

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Detailed Solution

Five numbers are selected from x1,x2x3,.....,x50

Total number ways of selecting 5 numbers=n(S)=50C5

x20as the middle number i
n(E)=30C2×19C2=

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