Question

In how -many ways can 21 identical white balls and 19 identical black balls be arranged in a row so that, no two black balls be together 7 What will be the result if all the balls are considered to be different ?

Answer 1

There will be only one way of arranging all the 21 white identical bails and there will be 22 gaps in which we have to place 19 identical black balls. Hence we have only to. select 19 places out of these 22. Therefore the required number is ${}^{22}{C}_{19}={}^{22}{C}_3=\frac{22\times 21\times 20}{\mathrm{1.2.3}}=1540.$
2nd Case: If the balls were all different then these can be arranged, in 21! ways and there will be 22 gaps in which 19 different balls can be arranged in ${}^{22}{P}_{19}$ ways i.e.$\frac{22!}{3!}$ ways. Hence by fundamental theorem, the required number of ways is $\frac{1}{6}\left(21\right)!\left(22\right)!$