Selection & Combination Questions with Solutions — Set 4 (Q31-Q40)

This set contains Selection and Combination Questions with Solutions — Set 4 (Q31-Q40) 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.

31. In the annual function of a school every student gave a gift to every other student. If total number of gift distributed is 380 then
(i). How many students were present in the function?
(ii). How many gifts received by a particular student Amit ?

32. How many factors of N = 2⁷ × 3⁷ × 5⁷ × 7⁷ ends with two zeros?

33. (i). Find the total number of factors of 54000
(ii). Sum of all factors of 54000
(iii). Sum of all factors of 54000 that ends with 5.

34. There are 10 singers who is ready to perform in a show that is scheduled for two timings, before lunch & after lunch. If before lunch maximum 6 singers can perform while after lunch maximum 5 singers can perform then in how many different way program can be scheduled?

35. In how many ways 5 students from a group of 10 students be selected if Ankit & Rahul are not selected together?

36. In how many ways a team of 4 boys & 5 girls out of total 8 boys & 10 girls can be formed?

37. In how many ways 5 players can be selected out of 10 players
(i). If Ankit & Rahul are always selected
(ii). If Ankit & Rahul are always rejected.

38. Find the number of ways of selecting 11 players from a group of 35 iplayers such that the oldest & the youngest players are always included in the group.

39. In how many ways 5 vowels & 6 consonants can be selected from 26 letters of alphabet?

40. At an election three wards of a town are canvassed by 6, 7 & 10 men repectively. If there are 30 volunteers, in how many ways can they be allotted to different wards?

31. In the annual function of a school every student gave a gift to every other student. If total number of gift distributed is 380 then
(i). How many students were present in the function?
(ii). How many gifts received by a particular student Amit ?
Solution:-
Let there are total n students in school.
2 students can be selected in ⁿC₂ ways
& since every student gave gift to every other student
Hence total number off gifts distributed = 2×ⁿC₂ = 380
⟹ 2 × \(\frac{{n(n – 1)}}{2}\)  = 380
n(n-1) = 380
∴ n = 20
Hence (i). 20 students were present in the function      Answer
(ii)
A particular student Amit will receive gift from remaining all students except Amit = 19 Answer

32. How many factors of N = 2⁷ × 3⁷ × 5⁷ × 7⁷ ends with two zeros ?
Solution:-

In every factor 2² × 5² will be common
∴ required number of factor is same as the number of factors of 2⁵ × 3⁷ × 5⁵ × 7⁷
= 6×8×6×8
= 2304   Answer

33. (i). Find the total number of factors of 54000
(ii). Sum of all factors of 54000
(iii). Sum of all factors of 54000 that ends with 5.
Solution:-
5400 = 2⁴ × 3³ × 5³

(i).
∴ total number of factors = (4 + 1)(3 + 1)(3 + 1)
= 80 Answer

(ii)
Sum of all the factors   =  \(\frac{{{2^{4 + 1}} – 1}}{{2 – 1}} \times \frac{{{3^{3 + 1}} – 1}}{{3 – 1}} \times \frac{{{5^{3 + 1}} – 1}}{{3 – 1}}\)
= 31×40×312
= 386880   Answer

(iii).
its factor will end with 5 sonly when it is odd factor ⇒ all the factors of 3³ × 5³ will satisfy the condition. Hence sum of all the factors of 3³ × 5³ is given by
\(\frac{{{3^{3 + 1}} – 1}}{{3 – 1}} \times \frac{{{5^{3 + 1}} – 1}}{{5 – 1}}\)
= 40×156
∴ Required answer = 6240 Answer

34. There are 10 singers who is ready to perform in a show that is scheduled for two timings, before lunch & after lunch. If before lunch maximum 6 singers can perform while after lunch maximum 5 singers can perform then in how many different way program can be scheduled?
Solution:-
Here two cases will be formed:-

case(i).
Before lunch 6 & after lunch 5 then number of ways is ¹⁰C₆ = 210 ways (∵ we just have to select only 6 out of 10 before lunch remaining 4 will be automatically selected in 1 way)

Case(ii).
After lunch 5 then before lunch it should be 5 singers & that can be done in ¹⁰C₅ ways = 252 ways
Hence total number of ways is = 210 + 252
= 462 Answer

35. In how many ways 5 students from a group of 10 students be selected if Ankit & Rahul are not selected together?
Solution:-

number of ways to select 5 student from a group of 10 students without any restriction = ¹⁰C₅ ways
& when both Ankit & Rahul are selected then number of ways to select 5 student is = ⁸C₃
Hence required number of ways = ¹⁰C₅ – ⁸C₃ Answer 

36. In how many ways a team of 4 boys & 5 girls out of total 8 boys & 10 girls can be formed?
Solution:-

⁸C₄ × ¹⁰C₅ ways Answer

37. In how many ways 5 players can be selected out of 10 players
(i). If Ankit & Rahul are always selected
(ii). If Ankit & Rahul are always rejected.
Solution:-

(i). ⁸C₃ ways
(ii). ⁸C₅ ways    Answer

38. Find the number of ways of selecting 11 players from a group of 35 iplayers such that the oldest & the youngest players are always included in the group.
Solution:-

Since the oldest & the youngest players i.e. 2 particular players are always included in the group

Hence number of ways = ³⁵⁻²C₁₁₋₂
= ³³C₉ Answer

39. In how many ways 5 vowels & 6 consonants can be selected from 26 letters of alphabet?
Solution:-

vowels = a, e, i, o, u ⟹ total 5
∴ 5 vowels can be selected only 1 way
& 6 consonants can be selected from 21 consonants in ²¹C₆ ways
Hence total number of ways = 1ײ¹C₆ = ²¹C₆  Answer

40. At an election three wards of a town are canvassed by 6, 7 & 10 men repectively. If there are 30 volunteers, in how many ways can they be allotted to different wards?
Solution:-

1st we will select 6 out of 30 in ³⁰C₆ ways then next 7 out of 24 in ²⁴C₇ ways & finally 10 in ¹⁷C₁₀ ways.

Hence total number of ways is = ³⁰C₆ײ⁴C₇×¹⁷C₁₀ Answer

✅ Well done on completing Set 4!

Continue practising with Selection and Combination Questions 41 to 50 → Set 5 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.