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:

In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

A) 360 B) 700
C) 720 D) 120
 
Answer & Explanation Answer: C) 720

Explanation:

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

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49 48964
Q:

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A) 120960 B) 120000
C) 146700 D) None of these
 
Answer & Explanation Answer: A) 120960

Explanation:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

 

Thus, we have MTHMTCS (AEAI).

 

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

 

Number of ways of arranging these letters = 8!/(2! x 2!)= 10080.

 

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

 

Number of ways of arranging these letters =4!/2!= 12.

 

Required number of words = (10080 x 12) = 120960

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3 8628
Q:

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

A) 4050 B) 3600
C) 1200 D) 5040
 
Answer & Explanation Answer: D) 5040

Explanation:

'LOGARITHMS' contains 10 different letters.

 

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

 

10P4

 

= 5040.

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43 42415
Q:

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

A) 54 B) 64
C) 63 D) 36
 
Answer & Explanation Answer: C) 63

Explanation:

Required number of ways = 7C5 × 3C2=63

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0 4366
Q:

In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

A) 36 B) 25
C) 42 D) 120
 
Answer & Explanation Answer: A) 36

Explanation:

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

 

Let us mark these positions as under: 

                                                      (1) (2) (3) (4) (5) (6) 

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.  

Number of ways of arranging the vowels = 3P3 = 3! = 6.

 

Also, the 3 consonants can be arranged at the remaining 3 positions. 

Number of ways of these arrangements = 3P3 = 3! = 6. 

Total number of ways = (6 x 6) = 36.

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12 29836
Q:

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

A) 48 B) 64
C) 63 D) 45
 
Answer & Explanation Answer: B) 64

Explanation:

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

 

Required number of ways=3C1*6C2+3C2*6C1+3C3 = (45 + 18 + 1) =64

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4 13099
Q:

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

A) 11670 B) 12000
C) 11760 D) 20050
 
Answer & Explanation Answer: C) 11760

Explanation:

Required number of ways= 8C5×10C6 = (8C3×10C4)= 11760.

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1 4050
Q:

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

A) 209 B) 290
C) 200 D) 208
 
Answer & Explanation Answer: A) 209

Explanation:

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys). 

 

Required number of ways = 6C1*4C3+6C2*4C2+6C3*4C1+6C4  

6C1*4C1+6C2*4C2+6C3*4C1+6C2 = 209.

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11 21696