How many different ways are there to divide 12 people into 2 groups of any sizes?

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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

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

Welcome to the next lesson in the Permutation and Combination series. We’ve come a long way, and we have a long way to go. Hope you’re having fun!

This lesson will tackle the following problem

In how many ways can n distinct objects be divided into r groups, whose sizes are known?

Here are a few examples:

In how many ways can we divide 5 different objects into two groups, of sizes 3 and 2?

In how many ways can we divide 10 different objects into three groups, of sizes 3, 2 and 5?

In how many ways can we divide 10 different objects into 5 pairs?

Note that dividing into groups of size 2 and 3 is equivalent to dividing into groups of size 3 and 2. That is, only the sizes matter, not the order of the groups. Similarly dividing 10 objects into three groups of sizes 3, 2 and 5 will be considered same as their division into groups of sizes 2, 3 and 5 or 5, 2 and 3.

Let’s begin with the first case. 5 objects, to be divided into two groups, of size 2 and 3.

Here are the five objects:

And, what we want is this:

Or, this:

How can we find the number of ways to do so?

Well, all you have to do is select 2 objects from the 5 and set them aside, forming a group. And, you don’t have to worry about the second group, as setting aside 2 objects results in 3 objects being left over, which will form the other group.

And, this selection can be done in 5C2 ways or \( \large \frac{5!}{2!.3!} \) ways.

Note that we could also have selected three objects first, leaving behind 2. This could be done in 5C3 ways, which is exactly the same as the previous answer.

Let’s take the second example now. In how many ways can we divide 10 different objects into three groups, of sizes 3, 2 and 5?

The method will be the same as used previously. First, we’ll select 3 objects out of 10, forming the first group. This can be done in 10C3 ways.

Next, from the remaining 7 objects, we’ll select 2 objects and form the second group, in 7C2 ways. And, the third group gets formed on its own, as there’ll be 3 objects left over.

The number of ways to perform both these tasks will be 10C3.7C2 (using the multiplication principle), which equals \( \large \frac{10!}{3!.2!.5!}\)

And what if a lot of 14 different objects had to be divided into 4 groups of sizes 2,3, 4 and 5?

You guessed it right – the answer will be 14C2.12C3.9C4 or \( \large \frac{14!}{2!.3!.4!.5!}\)

We can now arrive at a formula for the same. Let’s say we have n different objects, and we’ve to divide them into r groups of sizes a1, a2, a3, …, ar. Using the same logic above, the number of ways to do so will be \( \large \frac{n!}{a_1!.a_2!.a_3!…a_r!}\)

A small note. For now, the group sizes will be considered as distinct. That is,

a1 ≠ a2 ≠ a3 ≠ … ≠ ar

Things will change a bit, if one or more of the groups have the same size. I’ll talk about that in the next lesson.