Question

In how many ways can 7 students stand in a queue?

• A. $5040$
• B. $7!$
• C. $6!$
• D. (a) and (b) above.

Answer 1

The correct option is D (a) and (b) above.

We know that the number of ways of arranging "n" things taken "k" at a time, with or without repetition is ${}_k^{n}P=\frac{n!}{(n-k)!}$

In the above question we have to arrange $7$ students by taking all $7$ at a time.i.e $n=7$ and $k=7$.

$\Rightarrow {}_7^{7}P=\frac{7!}{(7-7)!}=\frac{7!}{0!}$

$\Rightarrow {}_7^{7}P=7!=5040.$

Is there any other alternate ways of solving this problem?
Yes! there is.
Let's check it out.
So, the number of ways $7$ students stand in queue is similar to number of ways of filling $7$ vacant places without repetition using $7$ different things.


First place can be filled in $7$ different ways.
Second place can be filled in $6$ ways using remaining $6$ things.
Similarly, $3^{rd},4^{th},5^{th},6^{th}$ and $7^{th}$ place can be filled in $5,4,3,2,1$ different ways respectively.

So, total number of ways is $=7\times 6\times 5\times 4\times 3\times 2\times 1=7!=5040.$