Question

In how many ways \[5\] boys and \[3\] girls can be seated in a row so that two girls are together?

Answer 1

Hint: This is a basic and elementary question involving the concept of permutation and combination
Some basic formulas:
$^n{\operatorname{P} _r} = \dfrac{{n!}}{{(n - r)!}}$ $
  n = natural\,\,number \\
  r = positive\,\,\operatorname{int} eger \\
 $
$^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
$n! = n(n - 1)(n - 2)......3.2.1$
factorial $n$ or $\,n!$
Product of first ‘n’ Natural numbers


Complete step by step solution:
In our above question
The seating arrangement can be follow as:
$ - B - B - B - B - B$
$5$ boys can sit in any of $5$ places represented as $B$ in above figure.
So, number of ways they can sit,
$ = {\,^5}{P_5}$
$ = \dfrac{{5!}}{{(5 - 5)!}}$
$ = \dfrac{{5!}}{{0!}}$
$ = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{1}$
  $ = 120$ ways
 $3$ girls can sit at any of $6$ places marked as – in above figure
So, number of ways they can sit
\[ = {\,^6}{P_3}\]
$ = \dfrac{{6!}}{{(6 - 3)!}}$
$ = \dfrac{{6 \times 5 \times 4 \times 3}}{{3!}}$
$ = 120$ways
Hence, total number of ways the can sit
$
   = 120 \times 120 \\
   = 14400\,ways \\
 $


Note: Hence, basic concept of permutation is used to solve questions which states that:
A permutation is a mathematical technique that determines the number of possible arrangements in a set where the order of the arrangements matters.