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:

How many different four letter words can be formed (the words need not be meaningful using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R?

A) 59 B) 56
C) 64 D) 55
 
Answer & Explanation Answer: A) 59

Explanation:

The first letter is E and the last one is R.

 

Therefore, one has to find two more letters from the remaining 11 letters.

 

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

 

The second and third positions can either have two different letters or have both the letters to be the same.

 

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

 

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

 

Total number of possibilities = 56 + 3 = 59

Report Error

View Answer Workspace Report Error Discuss

10 13641
Q:

From 5 consonants and 4 vowels, how many words can be formed using 3 consonants and 2 vowels ?

A) 7600 B) 7200
C) 6400 D) 3600
 
Answer & Explanation Answer: B) 7200

Explanation:

From 5 consonants, 3 consonants can be selected in 5C3 ways.

 

From 4 vowels, 2 vowels can be selected in 4C2 ways.

 

Now with every selection, number of ways of arranging 5 letters is 5P5ways.

 

Total number of words = 5C3*4C2*5P5

 

                                = 10x 6 x 5 x 4 x 3 x 2 x 1= 7200

Report Error

View Answer Workspace Report Error Discuss

4 13493
Q:

In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys ?

A) 36 B) 25
C) 24 D) 72
 
Answer & Explanation Answer: B) 25

Explanation:

The toys are different; The boxes are identical 

 

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways 

a. 2, 2, 1 

b. 3, 1, 1 

 

Case a. Number of ways of achieving the first option 2 - 2 - 1 

 

Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the remaining 3 can be selected in 3C2 ways and the last toy can be selected in 1C1 way. 

 

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2 

 

Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways 5C2*3C2= 15 ways

 

 

Case b. Number of ways of achieving the second option 3 - 1 - 1

 

Three toys out of the 5 can be selected in 5C3 ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

 

Therefore, total number of ways of getting the 3 - 1 - 1 option is 5C3 = 10 = 10 ways.

 

 

 

Total ways in which the 5 toys can be packed in 3 identical boxes

 

= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.

Report Error

View Answer Workspace Report Error Discuss

21 12609
Q:

How many necklace of 12 beads each can be made from 18 beads of different colours?

A) 18! B) 18! x 19!
C) 18!(6 x 24) D) 18! x 30
 
Answer & Explanation Answer: C) 18!(6 x 24)

Explanation:

Here clock-wise and anti-clockwise arrangements are same.

 

Hence total number of circular–permutations: 18P122*12 = 18!6*24

Report Error

View Answer Workspace Report Error Discuss

22 12569
Q:

There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can a team of 11 be selected so that the team contains at least 4 bowlers?

A) 1170 B) 1200
C) 720 D) 360
 
Answer & Explanation Answer: A) 1170

Explanation:

Possibilities     Bowlers      Batsmen         Number of ways

 

                         6               9

 

         1              4                7              (6C4*9C7)

 

         2              5                6              6C5*9C6

 

         3              6                5              6C6*9C5

 

6C4*9C7 = 15 x 36 = 540

 

6C5*9C6 = 6 x 84 = 504

 

 6C6*9C5= 1 x 126 = 126

 

Total = 1170

Report Error

View Answer Workspace Report Error Discuss

11 12022
Q:

In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together ?

A) 18000 B) 17280
C) 17829 D) 18270
 
Answer & Explanation Answer: B) 17280

Explanation:

Let 4 girls be one unit and now there are 6 units in all.

 

They can be arranged in 6! ways.

 

In each of these arrangements 4 girls can be arranged in 4! ways. 

 

Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280

Report Error

View Answer Workspace Report Error Discuss

4 11287
Q:

A coach must choose five starters from a team of 12 players. How many different ways can the coach choose the starters ?

A) 569 B) 729
C) 625 D) 769
 
Answer & Explanation Answer: B) 729

Explanation:

Choose 5 starters from a team of 12 players. Order is not important.

 

12C5= 729

Report Error

View Answer Workspace Report Error Discuss

3 11090
Q:

A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

A) 126 B) 120
C) 146 D) 156
 
Answer & Explanation Answer: A) 126

Explanation:

There are 8 students and the maximum capacity of the cars together is 9.


We may divide the 8 students as follows


Case I: 5 students in the first car and 3 in the second
Case II: 4 students in the first car and 4 in the second


Hence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways.


Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways.


Therefore, the total number of ways in which 8 students can travel is:
8C3+8C4=56 + 70= 126

Report Error

View Answer Workspace Report Error Discuss

5 10672