Permutaion or Arrangement Questions with Solutions  — Set 2 (Q11-Q29)

This set contains Permutation or Arrangement Questions with Solutions — Set 1 (Q11-Q19) 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.

11. In how many ways 15 students can be arranged in a row such that x is ahead of y and who in turn is ahead of z?

12. What is the total number of signals that can be made by using 6 flags of different colour when any number of them may be used?

13. In how many ways 5 distinct volume of chemistry & 7 distinct volume of botany books can be arranged on a bookshelf such that all the chemistry books & all the botany books are together ?

14. In how many ways 5 distinct volume of chemistry & 7 distinct volume of botany books can be arranged on a bookshelf such that all the chemistry books & all the botany books are together ?

15. In how many ways 15 students can be arranged in a row such that A is always between B and C ?

16. In how many ways 15 student can be arranged in a row such that p is always before q and r is always before s ?

17. In how many ways 15 students can be arranged in a row such that A is always ahead of B ?

18. In how many ways batting order of 11 players can be made out of 15 players
(i). If A & B are always rejected
(ii). If A & B are always rejected but C & D are always selected.
(iii). If two players can only play as wicket keeper.

19. If x is the number of ways in which 5 students can be arranged on 15 chairs in a straight line and y is the number of ways in which 5 students out of 15 students can be arranged on 5 chairs in a straight line. Then what is the relation between x & y ?

11. In how many ways 15 students can be arranged in a row such that x is ahead of y and who in turn is ahead of z ?
Sol:- we can arrange three students x, y & z in 3! = 6 ways in which only one way ⟹ xyz i.e. when x is ahead of y and who in turn is ahead of z is of our use.

Now without any restriction 15 students can be arranged in 15! ways.
& we know that every 6 only 1 case satisfy the given condition
Hence total number of ways = \(\frac{{15!}}{6}\)     Answer

Method (2):     _x_y_z_
we have selected x, y & z and placed them in the desired condition.
Now there are 4 spaces left for next student.
So Fourth student can be placed at these 4 places in 4 ways & this will increase one more space for next student to be placed.
∴ 5ᵗʰ student can go at 5 empty places in 5 ways & after placing this student one more space will be created for next student & this process will go on till all student ae placed.
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15ᵗʰ student can be placed in 15 ways.
Hence total number of ways to get desired answer
= 4×5×6×7×………….×15
=    \(\frac{{\left( {1 \times 2 \times 3} \right) \times \left( {4 \times 5 \times 6 \times 7 \times …………. \times 15} \right)}}{{(1 \times 2 \times 3)}}\)
=    \(\frac{{15!}}{{3!}} = \frac{{15!}}{6}\)      Answer

13. In how many ways batting order of 11 players can be made out of 15 players if Rohit is always selected ?
Sol:- Since Rohit is always selected so we are to select 10 players out of remaining 14 players which can be done in ¹⁴C₁₀ ways.
Now these 11 players can be arranged in 11! ways.
Hence total number of ways = ¹⁴C₁₀.11!             Answer

14. In how many ways  5 distinct volume of chemistry & 7 distinct volume of botany books can be arranged on a bookshelf such that all the chemistry books & all the botany books are together ?
Sol:- 

chemistry books within bunch 1 can be arranged in 5! ways. Botany books within bunch 2 can be arranged in 7! ways. Now these 2 bunches can be arranged in 2! ways.
Hence total number of ways = 5!.7!.2!          Answer

15. In how many ways 15 students can be arranged in a row such that A is always between B and C ?
Sol:- 15 students without any condition can be arranged in 15! ways.
In these 15! ways following combination of A, B & C will be :-

So out of 6 arrangement of A, B, C we get 2 favourable cases.
i.e.

16. In how many ways 15 student can be arranged in a row such that p is always before q and r is always before s ?
Sol:- p, q, r, s can be arranged in 4! = 24 ways.
out of these 24 ways favourable cases are:-

So out o 24 ways 6 favourable ways are there
i.e. 

17. In how many ways 15 students can be arranged in a row such that A is always ahead of B ?
Sol:- Without any condition 15 students can be arranged in 15! ways.
In these arrangements in 50% cases A will be ahead of B and in remaining 50% cases B will be ahead of A.
Hence number of ways when A is ahead of B =

18. In how many ways batting order of 11 players can be made out of 15 players
(i). If A & B are always rejected
(ii). If A & B are always rejected but C & D are always selected.
(iii). If two players can only play as wicket keeper.
Sol:-
(i). ∵ A & B are always rejected.
So we have to select 11 players out of remaining 15 – 2 = 13 players which can be done in ¹³C₁₁ ways
Now these 11 players can be arranged in 11! ways.
Hence total number of ways
= ¹³C₁₁.11!      Answer

(ii). A & B are always rejected then we have to select 11 out of 15 – 2 = 13
again C & D are always selected then we have to select 11 – 2 = 9 players out of 13 – 2 = 11 players which can be done in ¹¹C₉ ways.
Now these selected 11 players can be arranged in 11! ways.
Hence total number of ways
= ¹¹C₉.11!           Answer

(iii). From 2 wicket keeper only 1 is selected and this can be done in ²C₁ ways.
Now remaining 10 players can be selected out of 15 -2 = 13 players in ¹³C₁₀ ways.
So total number of ways of selection = ²C₁×¹³C₁₀
Now these 11 players can be arranged in 11! ways
Hence total number of ways = (²C₁×¹³C₁₀)(11!)        Answer

19. If x is the number of ways in which 5 students can be arranged on 15 chairs in a straight line and y is the number of ways in which 5 students out of 15 students can be arranged on 5 chairs in a straight line. Then what is the relation between x & y ?
Sol:-
 x = (number of ways to arrange 5 students on 15 chairs in a straight line)

y = (number of ways to arrange 5 students out of 15 students in 5 chairs)

Hence    x = y     Answer

✅ Well done on completing Set 2!

Now revisit the Permutation or Arrangement Concept Page to strengthen your formulas and tricks before moving ahead.

Consistent practice is the key to mastering percentage 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 Permutation on Wikipedia before attempting the next set.

This page is part of our complete series of permutation or arrangement questions 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 percentage concept before your exam day.