Permutations and Combinations 1

Hello learners, 

We are back after 6 years! Advance New Year wishes to all! 

Let's start with some Quantitative Aptitude topics! Today, let's begin with Permutations and Combinations - Day 1. 

Let's begin with the formulae! 

1. Permutation (Arrangement):

$$n{P}_r=\frac{n!}{(n-r)!}$$

Where $n$ is the total number of items, $r$ is the number of items to arrange, and $n! = n \times (n-1) \times \dots \times 1$.

2. Permutation (Arrangement):

$$n{P}_r=\frac{n!}{(n-r)!}$$

Where $n$ is the total number of items, $r$ is the number of items to arrange, and $n! = n \times (n-1) \times \dots \times 1$.

3. Factorial:

• $0! = 1$
• $n! = n \times (n-1) \times (n-2) \times \dots \times 1$

4. Repetition Permutation (when some items are identical):

$$\text{Total arrangements} = \frac{n!}{p_1! \times p_2! \times \dots \times p_k!}$$

Where $p_1, p_2, \dots, p_k$ are the frequencies of identical items.

5. Circular Permutation:

• For $n$ distinct items in a circle: $(n-1)!$
• If rotation is considered the same, divide further by $n$.

Question 1: Permutations

How many different ways can 4 people sit in a row?

Hint : Use nPr for arrangements

This is an arrangement problem, so use $n{P}_r=\frac{n!}{(n-r)!}$

Here, $n = 4$ and $r = 4$.

$$4P_4=\frac{4!}{(4-4)!}=\frac{4!}{0!}$$

Simplify the factorials.

$$4! = 4 \times 3 \times 2 \times 1 = 24,\quad 0! = 1$$
$$4P_4=\frac{24}{1}=24$$

Answer: There are 24 ways.

Question 2: Combinations

In how many ways can 3 students be selected from a group of 5 students?

Hint : Use nCr for combinations

This is a selection problem, so use $n{C}_r=\frac{n!}{r!\times (n-r)!}$

Here, $n = 5$and $r = 3$.

$$5C_3=\frac{5!}{3!\times (5-3)!}=\frac{5!}{3!\times 2!}$$

Simplify the factorials.

$$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120,\quad 3! = 3 \times 2 \times 1 = 6,\quad 2! = 2 \times 1 = 2$$
$$5C_3=\frac{120}{6\times 2}=\frac{120}{12}=10$$

Answer: There are 10 ways.

Question 3: Permutations with Repetition

How many unique arrangements can be made with the letters of the word SUCCESS?

Hint : Apply the repetition formula

This is a permutation problem with repetition, so use:

$$\text{Total arrangements}=\frac{n!}{p_1!\times p_2!\times \cdots \times p_k!}$$

Count the letters.

The word SUCCESS has 7 letters in total, with some letters repeating:

• S occurs 3 times.
• C occurs 2 times.
• U occurs 1 time.
• E occurs 1 time.

$$\text{Total arrangements}=\frac{7!}{\mathrm{3!}\times 2! \; \times \; 1! \; \times \; 1! }$$

Simplify the factorials.

7!=7×6×5×4×3×2×1=5040    3!=3×2×1=6    2!=2×1=2    1!=1

Answer: There are 420 unique arrangements of the letters in the word SUCCESS.

Question 4: Mixed (Selection + Arrangement)

A committee of 3 people is to be formed from a group of 6 people. In how many ways can the committee be formed if the positions are President, Secretary, and Treasurer?

Hint : Combine nCr​ for selection and r! for arrangement

This involves selecting and arranging, so use $n{P}_r=\frac{n!}{(n-r)!}$

Here, $n = 6$ and $r = 3$.

$$6P_3=\frac{6!}{(6-3)!}=\frac{6!}{3!}$$

Simplify the factorials.

$$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720,\quad 3! = 3 \times 2 \times 1 = 6$$
$$6P_3=\frac{720}{6}=120$$

Answer: There are 120 ways.

Question 5: Circular Permutation

In how many ways can 7 friends sit around a circular table?

Hint : Apply the circular permutation formula

For circular arrangements of $n$ items, use:

$$(n-1)!$$

Here, $n=\mathrm{7:}$

$$(7-1)!=6!$$

Simplify the factorial.

$$\mathrm{<span\; class="base"><span\; class="mord">6</span><span\; class="mclose">!</span><span\; class="mspace"></span><span\; class="mrel">=</span><span\; class="mord">6</span><span\; class="mspace"></span><span\; class="mbin">\times </span><span\; class="mord">5</span><span\; class="mspace"></span><span\; class="mbin">\times </span><span\; class="mord">4</span><span\; class="mspace"></span><span\; class="mbin">\times </span><span\; class="mord">3</span><span\; class="mspace"></span><span\; class="mbin">\times </span><span\; class="mord">2</span><span\; class="mspace"></span><span\; class="mbin">\times </span><span\; class="mord">1</span><span\; class="mspace"></span><span\; class="mrel">=</span>720</span>}$$

Answer: There are 720 ways.

Have fun with permutations and combinations—arranging and selecting in endless ways!