Find the number of ways of arrangement 6 red balls and 6 black balls in a row such that no two black ball…
Find the number of ways of arrangement 6 red balls and 6 black balls in a row such that no two black ball are together.
$6 ! \times 6 !$
$7 ! \times 6$ !
$2 \times 6 ! \times 6 !$
$7 \times 6 ! \times 6 !$
Solution
Number of ways to arrange 6 red balls are 6 !. Now, there are seven positions to place $6 \mathrm{black}$ balls, such that no two black balls are together
So, number of ways to arrange 6 black balls are ${ }^7 P_6$.
Therefore the required number of arrangements
$
=6 ! \times{ }^7 P_6=6 ! \times 7 !=7 \times 6 ! \times 6 !
$