Question

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

Answer 1

Hint: 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.