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Question

(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$

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Vedclass · Accepted Answer

$\frac{7}{{15}}$

Vedclass · Answer

(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 	imes 7 	imes 6}}{{1 	imes 2 	imes 3}} = 56$

So required probability $ = \frac{{56}}{{120}} = \frac{7}{{15}}.$

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