This set contains Selection and Combination Questions with Solutions — Set 9 (Q81-Q90) covering a mix of question types and difficulty levels — from basic to advanced — exactly as asked in real competitive exams.
Solutions are written in a simple, step-by-step notebook style for easy self-study and quick understanding. Each solution is broken down step by step so even the toughest question feels easy. These questions are hand-picked for students preparing for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests. International students preparing for GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests will find these equally useful.
✏️ Attempt each question on your own first — then check the solution below.
Selection & Combination Questions 81 to 90 with Solutions
81. In an election number of contestants is one more than the number of seats. If a voter can give minimum 1 and maximum equal to the number of seats vote then in how many different ways a particular voter can vote?
(i) 2044
(ii)2045
(iii) 2046
(iv) none of these
82. In the previous question to vote for 1 candidate a voter needs 1 ballot paper then which one of the following could be the total number of ballot paper:
(i). 4401
(ii). 4402
(iii). 4403
(iv). 4405
83. Consider a set s = {1, 2, 3, ………., 200}, two elements p and q are selected from this set s such that 7ᵖ + 7ᑫ is divisible by 5. In how many ways this selection can be done?
84. In an election the number of contestant is one more than the number of seats. A voter can give minimum 1 and maximum equal to the number of seats. Then find the number of contestant if a particular voter can vote in 510 different ways.
85. A committee of 12 is to be formed from 9 women and 8 men.
(i). In how many ways this can be done if at least 5 women have to be included in a committee?
(ii). In how many of these committees women are in majority?
(iii). In how many of these committees the men are in majority?
86. Let A and B be two sets containing 2 & 4 elements respectively. The number of subsets of A×B having 3 or more elements is?
87. In a certain test there are n questions. In this test 2ᵏ students gave wrong answers to at least n-k questions. k = 0, 1, 2, ……n. If the number of wrong answers is 8191 then what is the value of n?
88. A class contains 4 boys and x girls. In a test only 5 students including at least 3 boys can appear. If different group of students appear for the exam every time, if the number of times test conducted is 435 then find the total number of students in the class.
89. In a school, roll no. of k students are given from 1 to k. Three students are selected from these k students such that their roll numbers are not consecutive. In how many ways selection can be made?
90. A student is allowed to select at most n books from a collection of (2n+1) books. If the total number of ways in which he can select at least one book is 63, then what is the value of n?
Selection & Combination Questions 81 to 90 — Step-by-Step Solutions
81. In an election number of contestants is one more than the number of seats. If a voter can give minimum 1 and maximum equal to the number of seats vote then in how many different ways a particular voter can vote?
(i) 2044
(ii)2045
(iii) 2046
(iv) none of these
Solution:-
Let number of contestants is = n
∴ number of seats = n – 1
∴ number of ways a voter can vote is
= ⁿC₁ + ⁿC₂ + ⁿC₃ + ……………ⁿCₙ₋₁
= (ⁿC₀ + ⁿC₁ + ⁿC₂ + ⁿC₃ + ……………ⁿCₙ₋₁ + ⁿCₙ) – (ⁿC₀ +ⁿCₙ)
= 2ⁿ – 2
So total number of ways in which a voter can vote should be in 2ⁿ – 2 form, from the given options only 2046 is in this format.
Hence the Answer is (iii). 2046 Answer
82. In the previous question to vote for 1 candidate a voter needs 1 ballot paper then which one of the following could be the total number of ballot paper:
(i). 4401
(ii). 4402
(iii). 4403
(iv). 4405
Solution:-
From the answer of previous question a particular voter can vote in (2ⁿ – 2) ways
If number of voter is p
so total number of ways to vote = p×(2ⁿ – 2)
this will be same as the number of ballot paper required.
Hence number of ballot papers = p×(2ⁿ – 2)
From the given options
4402 = 71×62
= 71×(2⁶ – 2)
Hence Answer is option (ii). 4402
83. Consider a set s = {1, 2, 3, ………., 200}, two elements p and q are selected from this set s such that 7ᵖ + 7ᑫ is divisible by 5. In how many ways this selection can be done?
Solution:-
For a number to be divisible by 5, its last digit i.e. unit digit should be either 0 or 5 and unit digit of 7ⁿ can be 1, 3, 7 or 9
type of n Unit digit
7⁴ˣ 1
7⁴ˣ⁺¹ 7
7⁴ˣ⁺² 9
7⁴ˣ⁺³ 3
Hence from any combination of above unit digit taken two at a time last 5 can not be acheived but we can acheive last digit as 0 in the following way:-
p q
4x 4x + 2
4x + 2 4x
4x + 3 4x + 1
4x + 1 4x + 3
Here in set s = {1, 2, 3, 4, 5, ……………, 200}
Number of numbers of 4x type = 50
Number of number of 4x + 1 type = 50
Number of number of 4x + 2 type = 50
Number of number of 4x + 3 type = 50
∴ required number of ways of selection
= ⁵⁰C₁×⁵⁰C₁ + ⁵⁰C₁×⁵⁰C₁ + ⁵⁰C₁×⁵⁰C₁ + ⁵⁰C₁×⁵⁰C₁
= 4×⁵⁰C₁×⁵⁰C₁
= 4×50×50
= 10000 ways Answer
84. In an election the number of contestant is one more than the number of seats. A voter can give minimum 1 and maximum equal to the number of seats. Then find the number of contestant if a particular voter can vote in 510 different ways.
Solution:-
Let the number of contestants = n
∴ number of seats = n – 1
So a particular voter can give 1 or 2 or 3 or ……….. (n -1) votes.
∴ Number of ways a voter can vote is =
ⁿC₁ + ⁿC₂ + ⁿC₃ + …………….. + ⁿCₙ₋₁
=(ⁿC₀ + ⁿC₁ + ⁿC₂ + ⁿC₃ + …………….. + ⁿCₙ₋₁ + ⁿCₙ) – (ⁿC₀ + ⁿCₙ)
⟹ 2ⁿ – 2 = 510 (given in question)
⟹ 2ⁿ = 512
⟹ 2ⁿ = 2⁹
∴ n = 9
∴ Number of contestants = 9 Answer
85. A committee of 12 is to be formed from 9 women and 8 men.
(i). In how many ways this can be done if at least 5 women have to be included in a committee?
(ii). In how many of these committees women are in majority?
(iii). In how many of these committees the men are in majority?
Solution:-
(i).
5W7M + 6W6M + 7W5M + 8W4M + 9W3M
= ⁹C₅×⁸C₇ + ⁹C₆×⁸C₆ + ⁹C₇×⁸C₅ + ⁹C₈×⁸C₄ + ⁹C₉×⁸C₃
= 1008 + 2352 + 2016 + 630 + 56
= 6062 ways Answer
(ii).
When women are in majority then number of ways = ⁹C₇×⁸C₅ + ⁹C₈×⁸C₄ + ⁹C₉×⁸C₃
= 2016 + 630 + 56
= 2702 Answer
(iii).
When men are in majority then number of ways = ⁸C₇×⁹C₅ + ⁸C₈×⁹C₄
= 1008 + 126
= 1134 ways Answer
86. Let A and B be two sets containing 2 & 4 elements respectively. The number of subsets of A×B having 3 or more elements is?
Solution:-
A×B having will have 2×4 = 8 elements
∴ Required number of ways
= 2⁸ – ⁸C₀ – ⁸C₁ – ⁸C₂
= 256 – 1 – 8 – 28
= 219 Answer
87. In a certain test there are n questions. In this test 2ᵏ students gave wrong answers to at least n-k questions. k = 0, 1, 2, ……n. If the number of wrong answers is 8191 then what is the value of n?
Solution:-
Number of students who gave wrong answers to at lest r questions = 2ⁿ⁻ʳ
& Number of students who gave wrong answers to at least (r+1) questions = 2ⁿ⁻⁽ʳ⁺¹⁾
∴ Number of students who gave wrong answer to exactly r questions = 2ⁿ⁻ʳ – 2ⁿ⁻⁽ʳ⁺¹⁾
Then the total number of wrong answers is
= 1.(2ⁿ⁻¹ – 2ⁿ⁻²) + 2.(2ⁿ⁻² – 2ⁿ⁻³)+ 3.(2ⁿ⁻³ – 2ⁿ⁻⁴) + ………….. + r.(2ⁿ⁻ʳ – 2ⁿ⁻⁽ʳ⁺¹⁾) + …….. + n.(2⁰)
= 2ⁿ⁻¹ + 2ⁿ⁻² + 2ⁿ⁻³ + 2ⁿ⁻⁴ + …………….. + 2⁰
this above equation is sum of a G.P. which is equal to = 2ⁿ – 1
Hence 2ⁿ – 1 = 8191
2ⁿ = 8192
2ⁿ = 2¹³
∴ n = 13 Answer
88. A class contains 4 boys and x girls. In a test only 5 students including at least 3 boys can appear. If different group of students appear for the exam every time, if the number of times test conducted is 435 then find the total number of students in the class.
Solution:-
B G
4 x
Number of ways to select 5 student is
3B2G + 4B1G
= ⁴C₃ × ˣC₂ + ⁴C₄ × ˣC₁
= \(4 \times \frac{{x(x – 1)}}{2}\) + 1 × x
= 2x² – x
this is same as the number of test conducted which is 435
∴ 2x² – x = 435
2x² – x -435 = 0
2x²- 30x + 29x – 435 = 0
2x(x-15) + 29(x-15) = 0
(x-15)(2x+29) = 0
∴ x = 15 = number of girls in class
Hence total number of student in class
= 4 + 15
= 19 Answer
89. In a school, roll no. of k students are given from 1 to k. Three students are selected from these k students such that their roll numbers are not consecutive. In how many ways selection can be made?
Solution:-
roll numbers are:-
1, 2, 3, 4, 5, 6, 7, …………….., (k-2), (k-1), (k)
1ˢᵗ consider the number of ways when all the three roll number are consecutive. Then possible selections can be:-
(1, 2, 3) (2, 3, 4) (3, 4, 5) (4, 5, 6)…………….((k-2) (k-1)(k))
above selection of 3 consecutive roll numbers can be done in (k-2) ways
& number of ways of selection of three when roll number when there is no restriction = ᵏC₃
thus required number of ways = number of selection without any restriction – number of selection with restriction
= ᵏC₃ – (k-2)
= \(\frac{{k!}}{{3!(k – 3)!}}\) – (k-2)
= \(\frac{{k(k – 1)(k – 2)}}{6}\) – (k-2)
= \(\frac{{(k – 2)\{ {k^2} – k – 6\} }}{6}\)
= \(\boldsymbol{\frac{{(k – 2)(k – 3)(k + 2)}}{6}}\) Answer
90. A student is allowed to select at most n books from a collection of (2n+1) books. If the total number of ways in which he can select at least one book is 63, then what is the value of n?
Solution:-
total number of ways to select at least 1 & at most n books out of (2n+1) books is
= ²ⁿ⁺¹C₁ + ²ⁿ⁺¹C₂ + ²ⁿ⁺¹C₃ + ²ⁿ⁺¹C₄ + ……………… + ²ⁿ⁺¹Cₙ ……………….. equation (i)
= 63 (Given)
now ²ⁿ⁺¹C₀ + ²ⁿ⁺¹C₁ + ²ⁿ⁺¹C₂ + ²ⁿ⁺¹C₃ + ²ⁿ⁺¹C₄ +………….+ ²ⁿ⁺¹Cₙ₋₁ + ²ⁿ⁺¹Cₙ + ²ⁿ⁺¹Cₙ₊₁ + ²ⁿ⁺¹Cₙ₊₂ + ²ⁿ⁺¹Cₙ₊₃ + ………………. + ²ⁿ⁺¹C₂ₙ + ²ⁿ⁺¹C₂ₙ₊₁
= 2²ⁿ⁺¹ …………………… equation(ii)
equation (ii) can again be written as
²ⁿ⁺¹C₀ + ²ⁿ⁺¹C₂ₙ₊₁ + (²ⁿ⁺¹C₁ + ²ⁿ⁺¹C₂ + ²ⁿ⁺¹C₃ + ²ⁿ⁺¹C₄ +………….+ ²ⁿ⁺¹Cₙ₋₁ + ²ⁿ⁺¹Cₙ) + (²ⁿ⁺¹Cₙ + ²ⁿ⁺¹Cₙ₋₁ + ²ⁿ⁺¹Cₙ₋₂ + ……………….. + ²ⁿ⁺¹C₁)
= 2²ⁿ⁺¹
⟹ 1 + 1 +2(²ⁿ⁺¹C₁ + ²ⁿ⁺¹C₂ + ²ⁿ⁺¹C₃ +………….+ ²ⁿ⁺¹Cₙ) = 2²ⁿ⁺¹
⟹ 2 + 2(63) = 2²ⁿ⁺¹
⟹ 1+ 63 = 2²ⁿ
⟹ 2⁶ = 2²ⁿ
∴ n = 3 Answer
✅ Well done on completing Set 9!
Continue practising with Selection and Combination Questions 91 to 100 → Set 10 or revisit the Selection and Combination Concept Page to strengthen your formulas and tricks before moving ahead.
Consistent practice is the key to mastering Selection and Combination for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and international exams including GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests. Want to understand the concept better? Read about Combination (Mathematics) on Wikipedia before attempting the next set.
This page is part of our complete series of Selection and Combination Question with solutions for competitive exams — covering every question type from basic to advanced so you can build speed, accuracy and confidence. Practising these questions regularly will also strengthen your core LCM and HCF concept before your exam day.
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