Permutations and Combinations Questions

FACTS  AND  FORMULAE  FOR  PERMUTATIONS  AND  COMBINATIONS  QUESTIONS

 

 

1.  Factorial Notation: Let n be a positive integer. Then, factorial n, denoted n! is defined as: n!=n(n - 1)(n - 2) ... 3.2.1.

Examples : We define 0! = 1.

4! = (4 x 3 x 2 x 1) = 24.

5! = (5 x 4 x 3 x 2 x 1) = 120.

 

2.  Permutations: The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Ex1 : All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).

Ex2 : All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)

Number of Permutations: Number of all permutations of n things, taken r at a time, is given by:

Prn=nn-1n-2....n-r+1=n!n-r!

 

Ex : (i) P26=6×5=30   (ii) P37=7×6×5=210

Cor. number of all permutations of n things, taken all at a time = n!.

Important Result: If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,

such that p1+p2+...+pr=n

Then, number of permutations of these n objects is :

n!(p1!)×(p2! ).... (pr!)

 

3.  Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

Ex.1 : Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.

Note that AB and BA represent the same selection.

Ex.2 : All the combinations formed by a, b, c taking ab, bc, ca.

Ex.3 : The only combination that can be formed of three letters a, b, c taken all at a time is abc.

Ex.4 : Various groups of 2 out of four persons A, B, C, D are : AB, AC, AD, BC, BD, CD.

Ex.5 : Note that ab ba are two different permutations but they represent the same combination.

Number of Combinations: The number of all combinations of n things, taken r at a time is:

Crn=n!(r !)(n-r)!=nn-1n-2....to r factorsr!

 

Note : (i)Cnn=1 and C0n =1     (ii)Crn=C(n-r)n

 

Examples : (i) C411=11×10×9×84×3×2×1=330      (ii)C1316=C(16-13)16=C316=560

Q:

Find the number of ways to arrange 4 people in groups of 3 at a time where order matters?

A) 20 B) 16
C) 24 D) 36
 
Answer & Explanation Answer: C) 24

Explanation:

P(4,3)= P34= 24

Report Error

View Answer Report Error Discuss

0 3848
Q:

Find the number of ways to take 4 people and place them in groups of 3 at a time where order does not matter?

A) 4 B) 12
C) 36 D) 16
 
Answer & Explanation Answer: A) 4

Explanation:

Since order does not matter, use the combination formula 

C34 = 24/6 = 4

Report Error

View Answer Report Error Discuss

1 3748
Q:

How many more words can be formed by using the letters of the given word 'CREATIVITY'?

A) 851250 B) 751210
C) 362880 D) 907200
 
Answer & Explanation Answer: D) 907200

Explanation:

The number of letters in the given word CREATIVITY = 10

 

Here T & I letters are repeated

 

=> Number of Words that can be formed from CREATIVITY = 10!/2!x2! = 3628800/4 = 907200

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

8 3690
Q:

If A1, A2, A3, A4, ..... A10 are speakers for a meeting and A1 always speaks after, A2 then the number of ways they can speak in the meeting is 

A) 9! B) 9!/2
C) 10! D) 10!/2
 
Answer & Explanation Answer: D) 10!/2

Explanation:

As A1 speaks always after A2, they can speak only in  1st  to 9th places and 

 

A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place 

 

A2 can speak in 9 places the remaining 

 

 A3, A4, A5,...A10  has no restriction. So, they can speak in 9.8! ways. i.e

 

when A2 speaks in the first place, the number of ways they can speak is 9.8!.

 

When A2 speaks in second place, the number of ways they can speak is  8.8!.

 

When A2 speaks in third place, the number of ways they can speak is  7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!

 

 

 

Therefore,Total Number of ways they can  speak = (9+8+7+6+5+4+3+2+1) 8! = 92(9+1)8! = 10!/2

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

5 3646
Q:

If (1 × 2 × 3 × 4 ........ × n) = n!, then 15! - 14! - 13! is equal to ___?

A) 14 × 13 × 13! B) 15 × 14 × 14!
C) 14 × 12 × 12! D) 15 × 13 × 13!
 
Answer & Explanation Answer: D) 15 × 13 × 13!

Explanation:

15! - 14! - 13!

= (15 × 14 × 13!) - (14 × 13!) - (13!)

= 13! (15 × 14 - 14 - 1)

= 13! (15 × 14 - 15)

= 13! x 15 (14 - 1)

= 15 × 13 × 13!

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

9 3628
Q:

A,B,C,D,E,F,G and H are sitting around a circular table facing the centre but not necessarily in the same order. G sits third to the right of C. E is second to the right of G and 4th to the right of H. B is fourth to the right of C. D is not an immediate neighbour of E. A and C are immediate neighbours.

 Which of the following is/are correct ?

A) F is third to the left of B B) F is second to the right of B
C) B is an immediate neighbour of D D) All of the above
 
Answer & Explanation Answer: B) F is second to the right of B

Explanation:

From the given information, the circular arrangement is

EXP.

Here F is second to the right of B and the remaning all are wrong.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

6 3617
Q:

In how many ways the word 'SCOOTY' can be arranged such that 'S' and 'Y' are always at two ends?

A) 720 B) 360
C) 120 D) 24
 
Answer & Explanation Answer: D) 24

Explanation:

Given word is SCOOTY

ATQ,

Except S & Y number of letters are 4(C 2O's T)

Hence, required number of arrangements = 4!/2! x 2! = 4!

= 4 x 3 x 2

= 24 ways.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Analyst , Bank Clerk , Bank PO

6 3587
Q:

There are 41 students in a class, number of girls is one more than number of guys. we need to form a team of four students. all four in the team cannot be from same gender. number of girls and guys in the team should NOT be equal. How many ways can such a team be made ?

A) 49450 B) 50540
C) 46587 D) 52487
 
Answer & Explanation Answer: B) 50540

Explanation:

Given G + B= 41 and B = G-1
Hence, G = 21 and B = 20
Now we have 2 options,
1G and 3M
(or)
3G and 1M
(2G and 2M or 0G and 4M or 4G and oM are not allowed),

 

Total : C121×C320+C321×C120= 50540 ways.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

2 3561