Combination – Solved Problems
Combination deals with the selection of objects where the order of selection does not matter. Combination problems are commonly used in probability, statistics, and counting techniques.
Key Formulas:
Combination of r objects from n objects:
nCr = n! / [r!(n − r)!]
Relation with permutation:
nPr = nCr × r!
Combination of r objects from n objects:
nCr = n! / [r!(n − r)!]
Relation with permutation:
nPr = nCr × r!
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Problem 1: Basic combination
Find the number of ways to select 3 objects from 7 objects.
Solution:
n = 7, r = 3
7C3 = 7! / (3! × 4!)
= (7 × 6 × 5) / (3 × 2 × 1)
= 35
n = 7, r = 3
7C3 = 7! / (3! × 4!)
= (7 × 6 × 5) / (3 × 2 × 1)
= 35
Problem 2: Selecting students
In how many ways can a committee of 4 students be selected from 10 students?
Solution:
Order does not matter, so we use combination.
10C4 = 210
Order does not matter, so we use combination.
10C4 = 210
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Problem 3: Selecting objects of same type
How many ways can 2 pens be selected from 5 identical pens?
Solution:
Since pens are identical, only one selection is possible.
Answer = 1
Since pens are identical, only one selection is possible.
Answer = 1
Problem 4: Selecting with condition
From 5 men and 3 women, how many committees of 3 persons can be formed containing at least 1 woman?
Solution:
Total ways to select 3 from 8 people = 8C3 = 56
Ways with no women (only men) = 5C3 = 10
Required ways = 56 − 10 = 46
Total ways to select 3 from 8 people = 8C3 = 56
Ways with no women (only men) = 5C3 = 10
Required ways = 56 − 10 = 46
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Problem 5: Selecting prizes
In how many ways can 3 identical prizes be distributed among 7 students?
Solution:
Since prizes are identical, only selection matters.
Number of ways = 7C3 = 35
Since prizes are identical, only selection matters.
Number of ways = 7C3 = 35
Problem 6: Choosing subjects
A student must choose 2 subjects out of 6 subjects. In how many ways can the subjects be chosen?
Solution:
6C2 = (6 × 5) / 2 = 15
6C2 = (6 × 5) / 2 = 15
Problem 7: Property of combination
Evaluate: 9C2 + 9C7
Solution:
Using property: nCr = nCn−r
9C7 = 9C2
So sum = 2 × 9C2 = 2 × 36 = 72
Using property: nCr = nCn−r
9C7 = 9C2
So sum = 2 × 9C2 = 2 × 36 = 72
Problem 8: Word problem
How many ways can a team of 5 players be selected from a group of 12 players?
Solution:
12C5 = 792
12C5 = 792
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Key Takeaways
- Order does not matter in combinations
- Used mainly for selection problems
- nCr = nCn−r is a useful identity
- Always check if order is important before choosing formula