Combination

Combination is a fundamental concept in mathematics that refers to selecting items from a group, where the order does not matter. It is widely used in probability, statistics, and real-life problem-solving. The combination formula helps determine how many ways a subset can be chosen from a larger set, with or without repetition. Unlike permutations, combinations focus solely on the selection of objects, not their arrangement. Mastering combinations is essential for competitive exams like JEE and other advanced-level assessments.

1.0Combination Meaning

The term combination refers to the number of ways in which a group of items can be chosen from a larger set without considering the order of the items.

For example, choosing 2 fruits from a basket of {apple, banana, cherry} results in the combinations:

• {apple, banana}
• {apple, cherry}
• {banana, cherry}

Note: {apple, banana} and {banana, apple} are considered the same combination.

2.0Combination Formula in Maths

Standard Combination Formula

If you are choosing r items from n, the number of combinations is given by: ${}^n{C}_r=\frac{n!}{r!(n-r)!}$

Where:

• n = total number of items
• r = number of items chosen
• ! = factorial (e.g., 5! = 5 × 4 × 3 × 2 × 1)

Combination with Repetition Formula

If repetitions are allowed, the formula becomes: ${}^n{C}_r$ $\text{with repitition}=\frac{(n+r-1)!}{r!(n-1)!}$

This is used when you're selecting items and each item can appear more than once.

3.0Solved Examples on Combination 

Example 1: From a group of 7 students, how many ways can a committee of 3 students be formed?

Solution: ${}^7{C}_3=\frac{7!}{3!(7-3)!}=\frac{7\times 6\times 5}{P3\times 2\times 1}=35$

Example 2: How many ways can you choose 3 fruits from 5 types, if repetition is allowed?

Solution: ${}^{5+3-1}{C}_3{=}^7{C}_3=\frac{7!}{3!(7-3)!}=35$

Example 3: A team of 4 players is to be formed from 6 boys and 5 girls. If the team must include at least 2 girls, how many such teams can be formed?

Solution:

Break into cases:

• 2 girls, 2 boys: ${}^5{C}_2{\times }^6{C}_2=10\times 15=150$
• 3 girls, 1 boy: ${}^5{C}_3{\times }^6{C}_1=10\times 6=60$
• 4 girls: ${}^5{C}_4=5$

Total = 150 + 60 + 5 = 215

Example 4: A committee of 5 is to be formed from 8 men and 4 women. How many ways can it be done if the committee must contain at least 2 women?

Solution:
We consider all cases where number of women ≥ 2:

• Case 1: 2 women, 3 men: ${=}^4{C}_2{\times }^8{C}_3=6\times 56=336$
• Case 2: 3 women, 2 men: ${=}^4{C}_3{\times }^8{C}_2=4\times 28=112$
• Case 3: 4 women, 1 man: ${=}^4{C}_4{\times }^8{C}_1=1\times 8=8$

Total = 336 + 112 + 8 = 456 ways

Example 5: In how many ways can 4 identical balls be placed into 3 distinct boxes, where a box can hold any number of balls?

Solution:
This is a combination with repetition problem:

We must solve for non-negative integers $x_1+x_2+x_3=4$

Formula:  ${=}^{n+r-1}{C}_r{=}^{3+4-1}{C}_4{=}^6{C}_4=15$

Example 6: Prove using combination logic that:  ${\sum }_{k=0}^n{}^n{C}_k=2^n$

Solution:

This identity tells us the sum of all combinations of selecting 0 to n items from n items is $2^n$.

• Total number of subsets of a set with n elements = $2^n$
• ${}^n{C}_0{+}^n{C}_1+....{+}^n{C}_n=2^n$

Proved

Example 7: How many 4-letter words can be formed using the letters of the word ‘ENGINEERING’ such that all vowels are included only once?

Solution:

Letters in 'ENGINEERING':

E, N, G, I, N, E, E, R, I, N, G

Vowels = E, I (E ×3, I ×2) → usable once each → choose 2 vowels

Consonants available: N, G, R (multiple copies) → choose 2 distinct consonants

Let’s assume vowels used once only: 2 vowels from {E, I} (max possible 2)

→ We must choose 2 consonants from N, G, R

Ways to choose 2 vowels = ${}^2{C}_2=1$

Ways to choose 2 consonants (Distinct) =${}^3{C}_2=3$

Ways to arrange 4 chosen letters = $4!=24$

Total = 1 × 3 × 24 = 72

Answer = 72 ways

Example 8: In how many ways can 5 identical balls be distributed among 3 boys so that each boy gets at least 1 ball?

Solution:

We solve: $x_1+x_2+x_3=5$ where, $x_i\ge 1$

Convert to standard form: Let $y_i=x_i-1\Rightarrow y_1+y_2+y_3=2$

No of Solutions = ${}^{2+3-1}{C}_2{=}^4{C}_2=6$

Answer = 6 ways

4.0Applications of Combinations

• Probability theory: Finding the number of possible outcomes.
• Statistics: Selecting samples from populations.
• Algebra: Solving binomial expansion problems.
• Real life: Group selections, team formation, lottery systems.

5.0Key Differences: Permutation vs Combination