Permutations & Combinations | Formulas, Concepts & Shortcuts

In this article, we’ll explore the foundational concepts of permutations and combinations, provide essential formulas, and offer valuable shortcut tricks to efficiently solve problems. Whether you’re preparing for competitive exams or just curious about these concepts, this guide will make learning permutations and combinations easier.

What Are Permutations and Combinations?

Understanding the Difference: Permutations vs. Combinations

Before diving into formulas, it’s crucial to understand the core difference between permutations and combinations:

• Permutations: When order matters, you’re dealing with permutations. The arrangement of objects in a specific order is essential here.
• Combinations: When order doesn’t matter, it’s a combination. Here, only the selection of items counts, not their arrangement.

Example:

If you need to select 2 objects from {A, B, C}, here’s how it works:

• Permutations (order matters): AB, BA, BC, CB, AC, CA.
• Combinations (order doesn’t matter): AB, BC, CA.

Key Permutations and Combinations Formulas

1. Permutation Formula: Arrangements Matter

The permutation formula calculates the number of ways to arrange ‘r’ items out of ‘n’ total items:

P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n – r)!}P(n,r)=(n−r)!n!​

This formula is used when the order of selection is important.

Example:

If you want to arrange 3 objects out of 5 (A, B, C, D, E), the number of permutations is:

P(5,3)=5!(5−3)!=5!2!=60P(5, 3) = \frac{5!}{(5 – 3)!} = \frac{5!}{2!} = 60P(5,3)=(5−3)!5!​=2!5!​=60

2. Permutation with Identical Items

When some items are identical, the formula changes to:

P=n!p1!⋅p2!⋅⋯⋅pk!P = \frac{n!}{p1! \cdot p2! \cdot \dots \cdot pk!}P=p1!⋅p2!⋅⋯⋅pk!n!​

Where $p1,p2,\dots ,pk$ are the counts of identical items.

3. Circular Permutations: Arrangements Around a Circle

If you’re arranging items in a circle (where rotations matter), the formula becomes:

$$\text{Circular Permutations}=(n-1)!$$

For non-rotational arrangements, divide by 2 if the order doesn’t matter between clockwise and counter-clockwise.

Example:

For 6 people sitting at a round table, the total arrangements are:

$$(6-1)!=5!=120$$

4. Combination Formula: Selections Without Order

In contrast to permutations, combinations calculate how many ways you can select ‘r’ items from a set of ‘n’ items, where the order doesn’t matter:

C(n,r)=n!r!(n−r)!C(n, r) = \frac{n!}{r!(n – r)!}C(n,r)=r!(n−r)!n!​

Example:

Selecting 3 people from a group of 5:

C(5,3)=5!3!(5−3)!=10C(5, 3) = \frac{5!}{3!(5 – 3)!} = 10C(5,3)=3!(5−3)!5!​=10

5. Repetition in Selections

When repetition is allowed, the number of ways to select r items from n distinct items is:

$$C(n+r-1,r)$$

This formula applies when you can choose the same item more than once.

Problem Solving with Permutations and Combinations

Problem 1: Arranging Letters in Words

Example 1: How many ways can you arrange the letters in the word “TIGER”?

The number of arrangements for a word with n distinct letters is simply n!.

For “TIGER,” which has 5 distinct letters:

$$5!=120\text{ ways}$$

Problem 2: Arranging 3 Items from 5 Distinct Items

Example 2: In how many ways can 3 items be arranged from 5 distinct items?

First, recognize that this is a permutation problem because the order matters:

P(5,3)=5!(5−3)!=60P(5, 3) = \frac{5!}{(5 – 3)!} = 60P(5,3)=(5−3)!5!​=60

Problem 3: Arrangement with Identical Items

Example 3: In how many ways can you arrange the letters of “BBC”?

Since there are identical items (2 B’s), we adjust the permutation formula:

3!2!=3 ways(BBC,BCB,CBB)\frac{3!}{2!} = 3 \text{ ways} (BBC, BCB, CBB)2!3!​=3 ways(BBC,BCB,CBB)

Problem 4: Selecting Students for a Scholarship

Example 4: In how many ways can two students be selected from 10 for a scholarship?

Since the order doesn’t matter, this is a combination problem:

C(10,2)=10!2!(10−2)!=45 waysC(10, 2) = \frac{10!}{2!(10 – 2)!} = 45 \text{ ways}C(10,2)=2!(10−2)!10!​=45 ways

Problem 5: Forming a Committee

Example 5: A committee of 1 man and 3 women is to be formed from a group of 3 men and 4 women. How many ways can this be done?

First, calculate the number of ways to select 1 man and 3 women:

$$C(3,1)=3\text{(ways to select 1 man)}$$ $$C(4,3)=4\text{(ways to select 3 women)}$$

Total ways = $3\times 4=12$ ways.

Problem 6: Seated Arrangement at a Round Table

Example 6: 6 people are to be seated at a round table, but Sam will not sit next to Suhana. How many different ways can they sit?

Total number of circular permutations = $(6-1)!=120$

When Sam and Suhana are seated together, treat them as a single unit:

$$2!\times 4!=48\text{ ways}$$

Thus, the required number of ways = $120-48=72$.

Conclusion

Understanding permutations and combinations is key to solving many types of probability and counting problems efficiently. By mastering these formulas and applying them to practice problems, you can significantly improve your problem-solving skills. If you’re preparing for competitive exams, these concepts will frequently appear.