Question

In how many ways can 6 persons stand in a queue?

Answer 1

Hint: Numbers of ways of arrangement is determined by permutation.
Following is a short note to understand permutation and factorial in general:
A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.
The number of permutation of $n$ different objects taken $r$ at a time, where \[0 < r \le n\]and the objects do not repeat is $n(n - 1)(n - 2)...(n - r + 1)$, which is denoted by $P(n,r).$
                                                                           $P(n,r) = \dfrac{{n!}}{{(n - r)!}}$
The notation $n!$represents the product of first $n$natural numbers,
 i.e. the product \[1 \times 2 \times 3 \times ... \times (n - 1) \times n = n!\].
$\begin{array}{l}
0! = 1\\
1! = 1\\
2! = 2 \times 1\\
3! = 3 \times 2 \times 1
\end{array}$

Complete step by step solution:
Step 1: Given that
Total numbers of person = 6
$ \Rightarrow n = 6$
Step 2: find ‘r’
Numbers of persons standing in queue = 6
$ \Rightarrow r = 6$
Step 3: find the numbers of ways or arrangement:
Numbers of ways 6 persons can stand in a queue = $P(n,r) = P(6,6)$
                                                                                              $P(n,r) = \dfrac{{n!}}{{(n - r)!}}$
                                                                                                           $\begin{array}{l}
 = \dfrac{{6!}}{{(6 - 6)!}}\\
 = \dfrac{{6!}}{{0!}}\\
 = 6!\\
 = 6 \times 5 \times 4 \times 3 \times 2 \times 1\\
 = 720
\end{array}$

Therefore 6 persons can stand in a queue in 720 ways.

Note:
$n! = n(n - 1)!$
The number of permutations of $n$different things, taken r at a time, where repetition is allowed, is $\mathop n\nolimits^r $.
The number of permutations of $n$objects taken all at the time, where repetition is not allowed is $n!$.
Alternated steps:
Total numbers of persons =6
The number of permutation of 6 persons taken all at the time can stand in a queue $ = 6!$
                                                                                                                                                      $ = 720$
The number of permutation of $n$objects taken all at the time, where $\mathop p\nolimits_1 $objects are of first kind, $\mathop p\nolimits_2 $objects are of second kind, …, $\mathop p\nolimits_k $objects are of \[\mathop k\nolimits^{th} \]kind and rest, if any, are all different is
                                                                           $\dfrac{{n!}}{{\mathop p\nolimits_1 !\mathop p\nolimits_2 !...\mathop p\nolimits_k !}}$