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:

Ex : (i)       (ii) 

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 :

 

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:

Note : (i)      (ii)

Examples : (i)             (ii)

Q:

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

A) 25200 B) 52000
C) 120 D) 24400
 
Answer & Explanation Answer: A) 25200

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= \inline {\color{Black}(7C_3{}\times 4C_2{})}

= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves

= 5!

= 120

Required number of ways = (210 x 120) = 25200.

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Q:

How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

A) 376 B) 375
C) 500 D) 673
 
Answer & Explanation Answer: A) 376

Explanation:

The smallest number in the series is 1000, a 4-digit number.

The largest number in the series is 4000, the only 4-digit number to start with 4. 

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.

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31 13182
Q:

A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included ?

A) 196 B) 186
C) 190 D) 200
 
Answer & Explanation Answer: B) 186

Explanation:

When at least 2 women are included.

The committee may consist of 3 women, 2 men : It can be done in\inline \fn_jvn {\color{Black} 4C_{3}\times 6C_{2}}  ways

or, 4 women, 1 man : It can be done in \inline \fn_jvn {\color{Black} 4C_{4}\times 6C_{1}} ways

or, 2 women, 3 men : It can be done in \inline \fn_jvn {\color{Black} 4C_{2}\times 6C_{3}} ways.

Total number of ways of forming the committees

= \inline \fn_jvn {\color{Black} 4C_{2}\times 6C_{3}+4C_{3}\times 6C_{2}+4C_{4}\times 6C_{1}}

= 6 x 20 + 4 x 15 + 1x 6

= 120 + 60 + 6 =186

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12 11031
Q:

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

A) 720 B) 520
C) 700 D) 750
 
Answer & Explanation Answer: A) 720

Explanation:

The word 'LEADING' has 7 different letters.

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

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.

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

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

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7 8905
Q:

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :

A) 601 B) 600
C) 603 D) 602
 
Answer & Explanation Answer: A) 601

Explanation:

If the word started with the letter A then the remaining 5 positions can be filled in  5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.

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