How many ways can a team of 5 person to be formed out?

In mathematics, permutation is known as the process of arranging a set in which all the members of a set are arranged into some series or order. The process of permuting is known as the rearranging of its components if the set is already arranged. Permutations take place, in more or less important ways, in almost every area of mathematics. They frequently appear when different commands on certain finite sets are considered.

What is a Combination?

A combination is an act of choosing items from a group, such that (not like permutation) the order of choice does not matter. In smaller cases, it is possible to count the number of combinations. Combination refers to the union of n things taken k at a time without repetition  In combination you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used.

Permutation Formula

In permutation r things are selected from a set of n things without any replacement. In this order of selection matter.

nPr = (n!) / (n-r)!

Here,

n = set size, the total number of items in the set

r = subset size , the number of items to be selected from the set

Combination Formula

In combination r things are selected from a set of n things and where the order of selection does not matter.

nCr = n!/(n−r)!r!

Here, 

n = Number of items in set

r = Number of items selected from the set

In how many ways can 10 people be divided into two groups of five people?

Solution:

The first group can be chosen in 10C5 = 252 ways. There is just 1 way of choosing the second and final group from the 5 people who now remain.

In the process described above, every possible way of dividing 10 people into 2 group of 5 people each has been counted 2! = 2 times.

So the number of ways of dividing a group of 10 people into 2 group of 5 people each

= 252⁄2

= 126

Similar Questions

Question 1: In how many different ways can 8 people be divided into two groups of four people.

Solution:

The first group can be chosen in 8 C 4 = 70 ways. There is just 1 way of choosing the second and final group from the 4 people who now remain.

In the process described above, every possible way of dividing 8 people into 2 group of 4 people each has been counted 2! = 2 times.

So the number of ways of dividing a group of 8 people into 2 group of 4 people each

= 70⁄2

= 35

Question 2: How many different ways can a group of 8 people be divided into 4 teams of 2 people each?

Solution:

The first team can be chosen in 8 C 2 = 28 ways. Having done that, there are 6 people left and the second team can be chosen from them in 6 C 2 = 15 ways. 

After that there are 4 people left and the third team can be chosen from them in 4 C 2 = 6 ways. There is just 1 way of choosing the fourth and final team from the 2 people who now remain.

In the process described above, every possible way of dividing 8 people into 4 teams of 2 people each has been counted 4! = 24 times.

So the number of ways of dividing a group of 8 people into 4 teams of 2 people each

= 28 * 15 * 6 / 24

= 105

Answer

Verified

Hint: This question is from permutation and combinations. In this question, we are going to use the formula \[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]. This is the formula for when we have to select r things from n things.
First, we will see how many things we have to select from the total. After that, we will solve the question using the above formula.

Complete step by step answer:
Let us solve this question.
This is a question of permutation and combinations.
Let us first know what permutations and combinations.
A number of permutations (order matters) of n things taken r at a time:
\[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]
A number of combinations (order does not matter) of n things taken r at a time:
\[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]
So, in this question, we have to find that in how many ways, we can select a committee of five members from a group of 10 people.
So, in this question, the process of choosing the persons does not matter which means the order of choice does not matter.
 So, we are going to use the formula for this question is \[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]
We are choosing 5 persons from 10 persons where order does not matter.

Then, the numbers of ways are \[\dfrac{10!}{(10-5)!5!}\]
As we know that \[n!=n\times (n-1)\times (n-2)\times (n-3)\times ........\times 2\times 1\]
Then, \[\dfrac{10!}{(10-5)!5!}\] can be written as
\[\dfrac{10!}{(10-5)!5!}=\dfrac{10!}{5!\times 5!}=\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{\left( 5\times 4\times 3\times 2\times 1 \right)\times \left( 5\times 4\times 3\times 2\times 1 \right)}\]
Which is also can be written as
\[\dfrac{10!}{(10-5)!5!}=\dfrac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}=\dfrac{10}{5}\times 9\times \dfrac{8}{4}\times 7\times \dfrac{6}{3\times 2}=2\times 9\times 2\times 7\times 1=252\]
Hence, \[\dfrac{10!}{(10-5)!5!}=252\]
Therefore, the number of ways of selecting a committee of 5 members from a group of 10 persons is 252.

Note:
For solving this type of question, we should have a better knowledge of permutation and combinations. And, also remember the formulas of permutation and combination. Make sure that no calculation mistakes have been done in solving the question. Otherwise, the solution will be wrong. For example, \[\dfrac{10!}{(10-5)!5!}=\dfrac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}=\dfrac{10}{5}\times 9\times \dfrac{8}{4}\times 7\times \dfrac{6}{3\times 2}=2\times 9\times 2\times 7\times 1=252\]
In the above, mistakes can be made during calculations.

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