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14.Probability
medium
Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals
A
$\frac{1}{2}$
B
$\frac{7}{{15}}$
C
$\frac{2}{{15}}$
D
$\frac{1}{3}$
(IIT-1998)
Solution
(b) The number of ways to arrange $7$ white an $3$ black balls in a row $ = \frac{{10\,!}}{{7\,\,!\,.\,3\,\,!}} = \frac{{10.9.8}}{{1.2.3}} = 120$
Numbers of blank places between $7$ balls are $6$.
There is $1$ place before first ball and $1$ place after last ball. Hence total number of places are $8$.
Hence $3$ black balls are arranged on these $8$ places so that no two black balls are together in number of ways
$ = {}^8{C_3} = \frac{{8 \times 7 \times 6}}{{1 \times 2 \times 3}} = 56$
So required probability $ = \frac{{56}}{{120}} = \frac{7}{{15}}.$
Standard 11
Mathematics
