Combination questions with solutions are given here to practice and to understand how and when to use the concept of combinations while solving a problem.
Also, try important permutation and combination questions for class 11.
In combinatorics, the combination is a way of selecting something from a given collection. For example, we have to form a team of 4 people from the given ten persons. In combination, we can determine different ways of selecting 4 persons from 10 persons.
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The formula for Combination: To find the number of ways to select r items from n items without repetition, we have the formula for combination as follows:
\(\begin{array}{l}^{n}C_{r}=\binom{n}{r}=\frac{n!}{r!(n-r)!}\end{array} \)
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Learn more about Permutation and Combination.
Video Lesson on Formulas for Combination
Combination Questions with Solution
Practice the following combination questions, using the formulas for combination.
Question 1:
If 18Cr = 18Cr + 2, find rC5.
Solution:
We know that nCr = nC n – r, applying this formula
18Cr = 18Cr + 2
⇒ 18C18 – r = 18Cr + 2
⇒ 18 – r = r + 2
⇒ 2r = 18 – 2
⇒ r = 16/2 = 8
Then,
Question 2:
If nCr : nCr + 1 = 1 : 2 and nCr + 1 : nCr + 2 = 2 : 3, find the value of n and r.
Solution:
Given,
nCr : nCr + 1 = 1 : 2
⇒ n – 3r – 2 = 0 ….(i)
Also, nCr + 1 : nCr + 2 = 2 : 3
⇒ 2n – 5r – 8 = 0 ….(ii)
Multiplying (i) by 2 on both sides subtracting it from (ii), we get
–5r + 6r – 8 + 4 = 0
⇒ r = 4 and n = 14.
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Some Important Results:
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Question 3:
In how many ways 5 students can be chosen from 12 students?
Solution:
The required number of ways = 12C5
Question 4:
There are 10 people at a party who shakes hands with each other. If each two of them shake hands with each other, how many handshakes happen at the party?
Solution:
When two people shake hands it is counted as one handshake.
∴ Total number of handshakes = 10C2 = 10!/(2! × 8!) = 45.
Question 5:
How many diagonals are there of a 12-sided polygon?
Solution:
A diagonal can be formed by joining two non-adjacent vertices.
Number of diagonals of a 12 sided polygon = number of line segment in a 12 sided polygon – 12 edges of the polygon
= 12C2 – 12 = 12!/(2! × 10!) – 12
= 66 – 12 = 54.
∴ there are 54 diagonals.
Also read:
- Permutation and Combination Formula
- Difference Between Permutation and Combination
- Sequence and Series
Questions 6:
How many ways a 5-member committee can be formed out of 10 people if two particular people must be included?
Solution:
Number of people to be selected if two particular people must be included = 5 – 2 = 3
Number of ways of selecting 3 out of (10 – 2) = 8 people = 8C3 = 8!/(3! × 5!) = 56.
∴ there are 56 ways of forming such a committee
Question 7:
How many triangles can be formed using 10 points in a plane out of which 4 are collinear?
Solution:
Given 10 points on a plane out of which 4 are collinear, then 6 points are non-collinear.
Number of triangles formed by using 6 non-collinear points = 6C3 = 20
Number of triangles formed by using 6 non-collinear points and one out of the 4 collinear points = 6C2 × 4C1 = 15 × 4 = 60
Number of triangles formed by using 6 non-collinear points and two out of the 4 collinear points = 6C1 × 4C2 = 6 × 6 = 36
Total number of triangles can be formed = 20 + 60 + 36 = 116.
Question 8:
There are 4 balls of colour red, green, yellow and blue. In how many ways 2 two balls can be selected such that one of them is red or blue?
Solution:
Number of ways two balls can be selected out of 4 balls = 4C2 = 6 ways
Number of ways two balls can be selected such that neither red nor blue ball gets selected = (4 – 2)C2 = 2C2 = 1
Number of ways two balls can be selected such that either a red or a blue ball gets selected = 6 – 1 = 5 ways.
Question 9:
A team of four has to be selected from 6 boys and 4 girls. How many different ways a team can be selected if at least one boy must be there in the team?
Solution:
Combination of a four-member team with at least one boy are:
{(BGGG), (BBGG), (BBBG), (BBBB)}
Number of ways one boy and three girls can be selected = 6C1 × 4C3 = 6 × 4 = 24
Number of ways two boys and two girls can be selected = 6C2 × 4C2 = 15 × 6 = 90
Number of ways three boys and one girl can be selected = 6C3 × 4C1 = 20 × 4 = 80
Number of ways four boys can be selected = 6C4 = 15
Total number of ways to form such a team = 24 + 90 + 80 + 15 = 209.
Question 10:
It is compulsory to answer 10 questions in an examination choosing at least 4 questions from each part A and part B. If there are 6 questions in part A and 7 questions in part B, in how many ways can 10 questions be attempted?
Solution:
Case I: 4 questions from part A and 6 questions from part B
Number of ways of choosing 4 questions from part A = 6C4 = 15
Number of ways of choosing 6 questions from part B = 7C6 = 7
Total number of ways = 15 × 7 = 105
Case II: 5 questions from part A and 5 questions from part B
Number of ways of choosing 5 questions from part A = 6C5 = 6
Number of ways of choosing 5 questions from part B = 7C5 = 21
Total number of ways = 6 × 21 = 126
Case II: 6 questions from part A and 4 questions from part B
Number of ways of choosing 6 questions from part A = 6C6 = 1
Number of ways of choosing 4 questions from part B = 7C4 = 35
Total number of ways = 1 × 35 = 35
Required number of ways = 105 + 126 + 35 = 266.
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Practice Questions on Combinations
1. If 2nCr = 2nCr + 2, find the value of r.
2. Prove that n.(n – 1)C(r – 1) = (n – r + 1).nC(r – 1) for all 1≤ r ≤ n.
3. How many diagonals are there of a 15-sided polygon?
4. How many ways a 6-member committee can be formed out of 12 people if two particular people must not be included?
5. A committee of 5 persons have to be formed out of 3 women and 6 men, such that there should be at most 3 women. How many ways can such a committee be formed?
6. How many sides are there of a convex polygon which have 44 diagonals?
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