There are $10$ candidates for an examination out of which $4$ are appearing in $M a t h e m a t i c s$ and remaining $6$ are appearing in different subjects.Let "k" be the number of ways they can be seated in a row so that no two $M a t h e m a t i c s$ candidates are together. Find the sum of digits of k ?

Open in App

Solution

This question can be solved by Gap method.
In this method first, arrange the remaining candidates. Here the remaining candidates $= 6$ .
Hence the remaining candidates can be arranged in $6!$ ways.
$x O x O x O x O x O x O x$
$x$ : places available for $m a t h e m a t i c s$ candidates
$O$ : places for others
There are seven places available for $m a t h e m a t i c s$ candidates so that no two mathematics candidates are together. Now, $4$ candidates can be placed in these $7$ places in $^{7} C_{4}$ ways.
Hence, the total number of ways = $6! .^{7} C_{4} = 720 \times 840 = 604800$
Sum of digit is 18

Suggest Corrections

Similar questions

10 men and 6 women are to be seated in a row so that no two women sit together. the number of ways they can be seated is

10

6 women and 5 men are to be seated in a row so that no 2 men can sit together. Number of ways they can be seated is

4

5

Join BYJU'S Learning Program