Question

How do you find the permutation or combination of \[{}^{7}{{P}_{4}}\]?

Answer 1

Hint: On a set, the permutation is defined as the arrangement or rearrangement of the elements of the set in any order. The arrangement can be linear but we can also rearrange these elements in any other order. While the combination is defined as the selection of some elements or the other from the set without taking care of the order in which they are arranged. But it should be noted that permutations and combinations are not the same.

Complete step by step solution:
We have to find the permutation of \[{}^{7}{{P}_{4}}\]
We can make out from \[{}^{7}{{P}_{4}}\] that there is a total of $7$ elements from which selections have to be made and the number of selections to be made is $4$ and we have to arrange these $4$ elements.
So , for first selection , we have $7$ choices , for second selection , we are left with 6 choices for third we have 5 and for fourth we have $4$ choices. So, in all we have \[7\times 6\times 5\times 4\] choices which is the permutation of \[{}^{7}{{P}_{4}}\]. But we can also find this by the formula of permutation which is
\[\begin{align}
  & {}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!} \\
 \end{align}\]
In this question , $n=7$
and $r=4$
We put the values of n and r in the above formula to get
\[\begin{align}
 & {}^{7}{{P}_{4}}=\dfrac{7!}{(7-4)!} \\
 & =\dfrac{7!}{3!} \\
 & =\dfrac{(7\times 6\times 5\times 4\times 3\times 2\times 1)}{(3\times 2\times 1)} \\
\end{align}\]
Now Cancelling out the common terms, we get
\[{}^{7}{{P}_{4}}=7\times 6\times 5\times 4\]

which is the answer for the permutation of \[{}^{7}{{P}_{4}}\].

Note:
Permutations are of great importance. It helps us in counting the total number of ways in which a list of things can be arranged. Suppose we want to arrange $10$ balls in order then there are $10!$ ways in which the order of balls can be arranged.