Permutation Combination Examples

# Permutation Combination Examples

Permutation is defined as the arrangement of a collection of things, be it part or whole of the set of object where the importance is given to the order of the arrangement. While computing the formula of the permutation as n distinct objects taken $r$ at a time, we come up with $nPr$ = $\frac{n!}{(n – r)!}$. Here, $n$ stands for the total number of objects taken for permutation and $r$ stands for the number of objects out of total $n$ objects selected to form the permutation.

Combination is defined as the arrangement of a collection of things, be it part or whole of the set of object where no importance is given to the order of the arrangement. While computing the formula of the combination as $n$ distinct objects taken $r$ at a time, we come up with $nCr$ = $\frac{n!}{r! \times (n – r)!}$. Here, $n$ stands for the total number of objects taken for combination and $r$ stands for the number of objects out of total $n$ objects selected to form the combination.

## Word Problems

Example 1:

Six boxes are there numbering from $1$ to $6$. Each box contains either a red ball or a blue ball. One of the boxes must contain blue ball and that box containing the blue ball should be consecutively numbered. Find the total number of ways in which it can be done

Solution:

The different cases are as follows:

Case 1: When exactly one box contains a blue ball

In this case one single blue ball can be placed in any of the six boxes present. So, there are six ways of doing it. Red balls can be filled up in the remaining five boxes. As, red balls are identical there is one way of doing it.

So, total number of ways would be $6 \times 1$ = $6$

Case 2: When exactly two boxes contain blue balls

In this case there are two balls and the boxes need to be consecutively numbered as per given information. So the boxes could be (box $1$ and box $2$) or (box $2$ and box $3$) or (box $3$ and box $4$) or (box $4$ and box $5$) or (box $5$ and box $6$). The total number of ways of doing it is $5$. Red balls can be filled up in the remaining four boxes. As, red balls are identical there is one way of doing it.

So, total number of ways would be $5 \times 1$ = $5$

Case 3: When exactly three boxes contain blue balls

In this case there are three balls and the boxes need to be consecutively numbered as per given information. So the boxes could be (box $1$, box $2$ and box $3$) or (box $2$, box $3$ and box $4$) or (box $3$, box $4$ and box $5$) or (box $4$, box $5$ and box $6$). The total number of ways of doing it is $4$. Red balls can be filled up in the remaining three boxes. As, red balls are identical there is one way of doing it.

So, total number of ways would be $4 \times 1$ = $4$

Case 4: When exactly four boxes contain blue balls

In this case there are four balls and the boxes need to be consecutively numbered as per given information. So the boxes could be (box $1$, box $2$, box $3$ and box $4$) or (box $2$, box $3$, box $4$ and box $5$) or (box $3$, box $4$, box $5$ and box $6$). The total number of ways of doing it is $3$. Red balls can be filled up in the remaining two boxes. As, red balls are identical there is one way of doing it.

So, total number of ways would be $3 \times 1$ = $3$

Case 5: When exactly six boxes contain blue balls

In this case there are six balls and the boxes need to be consecutively numbered as per given information. So the box could be (box $1$, box $2$, box $3$, box $4$, box $5$ and box $6$). The total number of ways of doing it is $1$. No box is left for filling up with red ball. So, total number of ways would be $1$

Hence, total number of ways would be $6 + 5 + 4 + 3 + 2 + 1$ = $21\ ways$
Example 2:

In how many ways can $6$ identical balls be distributed in $4$ different boxes if any box can contain any number of balls

Solution:

Applying the formula of distribution of $k$ identical balls into $n$ boxes is $(k + n - 1)\ C\ (n – 1)$. Substituting the value of $k$ = $6$ and $n$ = $4$, we get $(7 + 4 – 1)\ C\ (4 – 1)$ = $10\ C\ 3$ = $120$

Example 3:

A board meeting of a company is organized in a room for $20$ persons along the two sides of a table with $10$ chairs in each side. $5$ persons wants to sit on a particular side and $3$ persons wants to sit on the other side. In how many ways can they be seated?

Solution:

Let us take up each side of the table at a time. On one side $5$ persons need to be seated in $10$ chairs. This can be done in $^{10}P_5$ = $30240\ ways$

On the opposite side $3$ persons need $t$ be seated in $10$ chairs. It is done in $^{10}P_3$ = $720\ ways$

Remaining persons = $20 – 5 – 3$ = $12$

Remaining chairs = $20 – 5 – 3$ = $12$

Now, we arrange the $12$ persons in $12$ chairs. It is done in $12!$ =

Required number of ways = $^{10}P_5 \times ^{10}P_3 \times 12!$
Example 4:

In how many different ways can the letters of the word 'MATHEMATICS' be arranged such that the vowels must always come together?

Solution:

Number of letters in the word ‘MATHEMATICS' is $11$ letters. The vowels present are four in number, they are $A, E, A$ and $I$. The vowels are taken together and considered a single letter $MTHMTCS(AEAI)$

So, total number of letters is $8, 7$ for MTHMTCS and $1$ for $AEAI$. Out of these $8$ letters $M$ is repeated twice and $T$ is also twice repeated. Thus, number of ways to arrange these letters = $\frac{8!}{2! \times 2!}$ = $10080$

Next in the group of vowels $AEAI, A$ is repeated twice. Thus, number of ways to arrange it = $\frac{4!}{2!}$ = $12$

Hence, required number of ways = $10080 \times 12$ = $120960$