Question

What’s the value of $^{10}{{P}_{5}}$ and $^{9}{{C}_{3}}$?

Answer 1

In this problem, two questions are asked. The first question is related to permutation of a set is a loose arrangement of its members into sequence or linear order. Second question is related to combination, which is a way to order or arrange a set or number of things. Formula which is used for both question that is for first question we use this formula $^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}$and for second question we use this formula $^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$

Complete step-by-step solution:
A permutation is a collection of objects arranged in a specific order. The members or elements or sets are arranged in this diagram in a sequential order.
Formulation for permutation is given by:
The following is a formula for permutation of n objects for r selection of objects that is
$P(n,r)=\dfrac{n!}{(n-r)!}$
We have to find the value of $^{10}{{P}_{5}}$. Here, we can observe that n (no. of objects) is 10 and r is 5 (no. of objects selected)
Now, applying the above formula to find the permutation –
$P(10,5)=\dfrac{10!}{(10-5)!}$
Now, we have to expand the factorial. Firstly
$10!=10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$
$(10-5)!=5!=5\times 4\times 3\times 2\times 1$
Now, after substituting this formula we get:
$P(10,5)=\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1}$
$^{10}{{P}_{5}}=30240$
So, we can say that there are 30240 ways of selecting 5 objects out of 10 objects.
Now, we go for the second question which is related to combination.
We will use the formula which is given by
$^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$
Here, in this question we have $n=9$ and $r=3$ after substituting this formula we get:
$^{9}{{C}_{3}}=\dfrac{9!}{3!(9-3)!}$
After simplifying this further we get:
$^{9}{{C}_{3}}=\dfrac{9!}{3!6!}$
Now, we have to find the each factorial and substitute in above equation we get:
$9!=9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$
$3!=3\times 2\times 1$
$6!=6\times 5\times 4\times 3\times 2\times 1$
Now, we have to substitute this factorial value into above equation:
$^{9}{{C}_{3}}=\dfrac{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{\left( 3\times 2\times 1 \right)\left( 6\times 5\times 4\times 3\times 2\times 1 \right)}$
After simplifying this we get:
$^{9}{{C}_{3}}=84$
Therefore, we get the value of $^{10}{{P}_{5}}=30240$ as well as $^{9}{{C}_{3}}=84$.

Note: In case of combination as well as permutation by just plugging the number in the given formula we can get the required answer. Remember from our factorial lesson that $n!=n\times (n-1)\times (n-2)........\times 2\times 1$. In combination order doesn’t matter but in case of permutation order may matter.