Selection Questions with Solutions — Set 7 (Q61 to Q70)

This set contains Selection and Combination Questions with Solutions — Set 7 (Q61-Q70) 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.

61. Let p be the number of factors of 3000 & q be the number of factors of 4000 & r be the number of common factors of 3000 and 4000. Then find the ratio of (p-r):(q-r)

62. If ‘p’ is the number of ways in which three numbers in A.P. can be selected from 1, 2, 3, 4, …….., n. Then find the value of p if
(i). n is odd
(ii). n is even

63. There are infinite number of letters F, L and Y. In how many ways we can choose ‘n’ letters such that the word ‘FLY’ can not be formed?

64. Let E = {1, 2, 3, 4} and F = {1, 2} then the number of onto functions from E to F will be?

65. A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw.

66. A question paper is divided into two parts A & B and each part contains 5 questions. The number of ways in which a candidate can answer 6 questions selecting at least two questions from each part is:

67. Out of 8 men and 4 women a committee of 5 is to be formed. In how many ways this can be done:
(i). So as to include at least 1 woman in the committee
(ii). So as to include at least 1 woman in the committee & exclude at least 1 man from the committee.

68. In how many ways two numbers can be selected from 100, 101, 102, ……….199 such that when these selected numbers are multiplied then the product is divisible by 3?

69. In how many ways can a mixed double game in Tennis be arranged from 20 married couple, if no husband and wife play in the same game.

61. Let p be the number of factors of 3000 & q be the number of factors of 4000 & r be the number of common factors of 3000 and 4000. Then find the ratio of (p-r):(q-r)
Solution:-
3000 = 3×10×10×10 = 2³×3¹×5³
∴ p = (3+1)(1+1)(3+1)
= 32
& 4000 = 4×10×10×10 = 2⁵×5³
∴ q = (5+1)(3+1) = 24
H.C.F. of 3000 & 4000 is 1000 = 10×10×10 = 2³×5³
∴ r = (3+1)(3+1) = 16
∴ (p-r):(q-r) = (32-16):(24-16)
= 16:8 = 2:1 Answer

62. If ‘p’ is the number of ways in which three numbers in A.P. can be selected from 1, 2, 3, 4, …….., n. Then find the value of p if
(i). n is odd
(ii). n is even
Solution:-
Given numbers are :-
1, 2, 3, 4, ………., n
Let the three selected numbers in A.P. be a, b, c then
a + c = 2b
Since 2b is even
Hence a+c should also be even
& This is possible only when both a & c are even or both a & c are odd.

Case(i). when n is odd
Let n = 2m + 1
∴ Number of odd numbers in given series = \(\frac{{(2m + 1) + 1}}{2}\)
= m+1
& Number of even numbers in given series = \(\frac{{(2m + 1) – 1}}{2}\)
= m
∴ number of ways of selection of a & c from (m+1) odd integers = ᵐ⁺¹C₂
& number of ways of selection of a & c from m even integers = ᵐC₂
∴ required number of ways = ᵐ⁺¹C₂ × ᵐC₂
= \(\frac{{(m + 1)m}}{2} \times \frac{{m(m – 1)}}{2}\)
= m²
= \({\left( {\frac{{n – 1}}{2}} \right)^2}\)        (∵ n = 2m + 1)
= \(\boldsymbol{\frac{1}{4}{(n – 1)^2}}\)  Answer

case(ii). when n is even
we can assume n = 2m
∴ Number of even integers in given series = m
& Number of odd integers in given series = m
∴ Number of ways of selection of a & c from m odd integers = ᵐC₂
Hence total number of ways =  ᵐC₂ +  ᵐC₂
=  \(\frac{{m(m + 1)}}{2} + \frac{{m(m – 1)}}{2}\)
= m(m-1)
=  \(\frac{n}{2}\left( {\frac{n}{2} – 1} \right)\)   (∵ n = 2m)
=  \(\boldsymbol{\frac{{n(n – 2)}}{4}}\)     Answer

63. There are infinite number of letters F, L and Y. In how many ways we can choose ‘n’ letters such that the word ‘FLY’ can not be formed?
Solution:-
The word ‘FLY’ can not be formed if the ‘n’ selected letters do not contain at least one of F, L and Y.
when ‘n’ letters are selected from F or L

every box has 2 ways of filling (either F or L)
Hence Number of ways of selecting ‘n’ letters which are F or L = 2ⁿ including the case when all the ‘n’ letters are F or all are L ……………………….. case(i)

Similarly, Number of ways of selecting ‘n’ letter which are F or Y =  2ⁿ including the case when all the ‘n’ letters are F or all are Y………………………. case(ii)

& number of ways of selecting ‘n’ letters which are L or Y = 2ⁿ including the case when all the ‘n’ letters are L or all are Y……………………………..case(iii)

Similarly, number of ways of selecting ‘n’ letters when all are F, L and Y are 1ⁿ = 1 in each case

Thus required number of ways
= 2ⁿ + 2ⁿ + 2ⁿ – 1 – 1 – 1
= 3(2ⁿ – 1)          Answer

64. Let E = {1, 2, 3, 4} and F = {1, 2} then the number of onto functions from E to F will be?
Solution:-
In set E every element has 2 option to match with set F either 1 or 2.
Hence total number of functions from E to F is 2⁴ = 16
out of these 16 functions we find that only two functions f and g given by
f(x) = 1        for all x in E
& g(x) = 2        for all x in E
are onto
& remaining 16 – 2 = 14 function are onto.    Answer

65. A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw.
Solution:-
Since here in question it is not mentioned that ball are identical so we will assume all balls to be different.
Hence required number of ways = 1 black AND 2 other OR 2 blackAND 1 other OR 3black and 0 other
= ³C₁×⁶C₂ + ³C₂×⁶C₁ + ³C₃
= 45 + 18 + 1
= 64     Answer

66. A question paper is divided into two parts A & B and each part contains 5 questions. The number of ways in which a candidate can answer 6 questions selecting at least two questions from each part is:
Solution:-

∴ total number of ways = 50 + 100 + 50
                                          = 200 ways     Answer

67. Out of 8 men and 4 women a committee of 5 is to be formed. In how many ways this can be done:
(i). So as to include at least 1 woman in the committee
(ii). So as to include at least 1 woman in the comittee & exclude at least 1 man from the committee.
Solution:-
(i).

Method(1):-

Hence required number of ways
= 280 + 336 + 112 + 8
= 736    Answer

Method(2):

Total number of ways to select he committee without any restriction = ¹²C₅
total number of way when no woman is selected i.e. all 5 are men = ⁸C₅
Hence required number of ways = ¹²C₅ – ⁸C₅
                                                           = 736    Answer

(ii)

Here also possible cases are same as the cases in (i).

Hence number of ways = ⁴C₁×⁸C₄ + ⁴C₂×⁸C₃ + ⁴C₃×⁸C₂ + ⁸C₁×⁴C₄
                                           = 736      Answer

68. In how many ways two numbers can be selected from 100, 101, 102, ……… 199 such that when these selected numbers are multiplied then the product is divisible by 3?
Solution:-

Numbers divisible by 3 are:-

& remaining 100 – 33 = 67 numbers are not multiple of 3.
Since product of two numbers is divisible by 3, it is possible only when at least one of them is multiple of 3.
∴ required number of ways 
= if both the numbers are multiple of 3 + if only one number is multiple of 3
= ³³C₂ + ³³C₁×⁶⁷C₁    Answer

69. In how many ways can a mixed double game in Tennis be arranged from 20 married couples, if no husband and wife play in the same game.
Solution:-
2 men can be selected in ²⁰C₂ ways

and that can be done in ¹⁸C₂ ways
Now we  have 2 men & 2 women & they can be pair up in 2 ways
Hence total number of ways = ²⁰C₂×¹⁸C₂×2              Answer

✅ Well done on completing Set 7!

Continue practising with Selection and Combination Questions 71 to 80 → Set 8 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.