Question

The number of ways in which the letters of the word $BALLON$ can be arranged so that two $L^{\prime }s$ do not come together, is

Answer 1

There are in all seven letters in the word $BALLON$ in which $L$ occurs $2$ times and $O$ occurs $2$ times.

$\therefore$ The number of arrangements of the seven letters of the word $=\frac{7!}{2!\times 2!}=1260$

If two $L^{\prime }s$ always come together, taking them as one letter,

we have to arrange $6$ letters in which $O$ occurs $2$ times.

$\therefore$ The number of arrangements in which the two $L^{\prime }s$ come together

$=\frac{6!}{2!}=6\times 5\times 4\times 3=360$

Hence the required number of ways in which the two $L^{\prime }s$ do not come together

$=1260-360=900.$