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:

The number of ways in which 8 distinct toys can be distributed among 5 children?

A) 5P8 B) 5^8
C) 8P5 D) 8^5
 
Answer & Explanation Answer: B) 5^8

Explanation:

As the toys are distinct and not identical,

For each of the 8 toys, we have three choices as to which child will receive the toy. Therefore, there are 58 ways to distribute the toys.

 

Hence, it is 58 and not 85.

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16 15154
Q:

In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together?

A) 1440 B) 720
C) 2250 D) 3600
 
Answer & Explanation Answer: D) 3600

Explanation:

Given word is THERAPY.

Number of letters in the given word = 7

These 7 letters can be arranged in 7! ways.

Number of vowels in the given word = 2 (E, A)

The number of ways of arrangement in which vowels come together is 6! x 2! ways

 

Hence, the required number of ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together = 7! - (6! x 2!) ways = 5040 - 1440 = 3600 ways.

 

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11 13113
Q:

In how many different ways can the letters of the word 'HAPPYHOLI' be arranged?

A) 89,972 B) 90,720
C) 72,000 D) 81,000
 
Answer & Explanation Answer: B) 90,720

Explanation:

The given word HAPPYHOLI has 9 letters

These 9 letters can e arranged in 9! ways.

But here in the given word letters H & P are repeated twice each

Therefore, Number of ways these 9 letters can be arranged is 

9!2! x 2! = 9 x 8 x 7 x 6 x 5 x 4 x 32 = 90,720 ways.

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7 2305
Q:

How many words can be formed with or without meaning by using three letters out of k, l, m, n, o without repetition of alphabets.

A) 60 B) 120
C) 240 D) 30
 
Answer & Explanation Answer: A) 60

Explanation:

Given letters are k, l, m, n, o = 5

number of letters to be in the words = 3

Total number of words that can be formed from these 5 letters taken 3 at a time without repetation of letters = 5P3 ways.

 5P3 = 5 x 4 x 3 = 60 words.

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9 3368
Q:

The letters of the word PROMISE are to be arranged so that three vowels should not come together. Find the number of ways of arrangements?

A) 4320 B) 4694
C) 4957 D) 4871
 
Answer & Explanation Answer: A) 4320

Explanation:

Given Word is PROMISE.

Number of letters in the word PROMISE = 7

Number of ways 7 letters can be arranged = 7! ways

Number of Vowels in word PROMISE = 3 (O, I, E)

Number of ways the vowels can be arranged that 3 Vowels come together = 5! x 3! ways

 

Now, the number of ways of arrangements so that three vowels should not come together

= 7! - (5! x 3!) ways = 5040 - 720 = 4320.

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7 6250
Q:

In how many different ways can the letters of the word 'POVERTY' be arranged ?

A) 2520 B) 5040
C) 1260 D) None
 
Answer & Explanation Answer: B) 5040

Explanation:

The 7 letters word 'POVERTY' be arranged in 7P7 ways = 7! = 5040 ways.

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13 2893
Q:

A decision committee of 5 members is to be formed out of 4 Actors, 3 Directors and 2 Producers. In how many ways a committee of 2 Actors, 2 Directors and 1 Producer can be formed ?

A) 18 B) 24
C) 36 D) 32
 
Answer & Explanation Answer: C) 36

Explanation:

Required Number of ways = 4C2 × 3C2 × 2C1 = 36 = 36

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17 3012
Q:

How many six digit odd numbers can be formed from the digits 0, 2, 3, 5, 6, 7, 8, and 9 (repetition not allowed)?

A) 8640 B) 720
C) 3620 D) 4512
 
Answer & Explanation Answer: A) 8640

Explanation:

Let the 6 digits of the required 6 digit number be abcdef

Then, the number to be odd number the last digit must be odd digit i.e 3, 5, 7 or 9

The first digit cannot be ‘0’ => possible digits = 3, 5, 7, 2, 6, 8

Remaining 4 places can be of 6 x 5 x 4 x 3 ways

This can be easily understood by

Therefore, required number of ways = 6 x 6 x 5 x 4 x 3 x 4 = 36 x 20 x 12 = 720 x 12

8640 ways.

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10 5195