Question

You are given $8$ balls of different colours (black, white, ...). The number of ways in which these balls can be arranged in a row so that the two balls of particular colour (say red and white) may never come together, is

• A. $8!-2\times 7!$
• B. $6\times 7!$
• C. $2\times 6!\times {}^7{C}_2$
• D. $2(7!\times 8!)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $8!-2\times 7!$
B $6\times 7!$
C $2\times 6!\times {}^7{C}_2$

Total number of permutations $=8!$
Let us take the two particular ball together and consider them as one.
Number of ways in which two particular balls come together $=2!\times 7!$ since two balls can also be permuted in $2!$ ways.
Required number of ways
$=8!-2\cdot (7!)=7!(8-2)=6\cdot 7!=6!\times 7\times 6=2\times 6!\times {}^7{C}_2$