In how many ways can 21 white balls and 19 bl | vedclass

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(B) To arrange $21$ white balls and $19$ black balls such that no two black balls are together,we use the gap method.First,arrange the $21$ white ball

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$1540$

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(B) To arrange $21$ white balls and $19$ black balls such that no two black balls are together,we use the gap method.First,arrange the $21$ white balls in a row. This creates $22$ possible gaps (including the ends) where the black balls can be placed.Number of ways to arrange $21$ white balls $= 1$.Number of ways to choose $19$ gaps out of $22$ for the black balls $= ^{22}C_{19}$.Using the property $^{n}C_{r} = ^{n}C_{n-r}$,we have $^{22}C_{19} = ^{22}C_{3}$.$^{22}C_{3} = \frac{22 	imes 21 	imes 20}{3 	imes 2 	imes 1} = 22 	imes 7 	imes 10 = 1540$.Total arrangements $= 1540 	imes 1 = 1540$.

In how many ways can $21$ white balls and $19$ black balls be arranged in a row such that no two black balls are together?

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