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 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 4173
Q:

How many such pair of letters are there in the word ‘TROUBLED’ which have as many letters between them in the word as they have between them in the English alphabet?

A) 2 B) 3
C) 4 D) 5
 
Answer & Explanation Answer: A) 2

Explanation:

Hence there are two pairs

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13 4171
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.

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2 4159
Q:

In how many ways can 4 sit at a round table for a group discussions?

A) 9 B) 3
C) 6 D) 12
 
Answer & Explanation Answer: C) 6

Explanation:

We get this using the formula of circular permutations i.e., (4 - 1)! = 3! =6

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0 4158
Q:

If each vowel of the word NICELY is changed to the next letter in the English alphabetical series and each consonant is changed to the previous letter in the English alphabetical series and then the alphabets are arranged in alphabetical order, which of the following will be fourth from the left?

A) M B) J
C) B D) O
 
Answer & Explanation Answer:

Explanation:

Thus after arranging the letters as per English alphabetical series; we get; Thus 4th letter from the left end will be K.

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16 4006
Q:

In how many ways can letter of the word RAILINGS arrange so that R and S always come together?

A) 1260 B) 2520
C) 5040 D) 1080
 
Answer & Explanation Answer: C) 5040

Explanation:

The number of ways in which the letters of the word RAILINGS can be arranged such that R & S always come together is

 

Count R & S as only 1 space or letter so that RS or SR can be arranged => 7! x 2!

 

But in the word RAILINGS, I repeated for 2 times => 7! x 2!/2! = 7! ways = 5040 ways.

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7 3996
Q:

From a group of 7 boys and 6 girls, five persons are to be selected to form a team, so that at least 3 girls are there in the team. In how many ways can it be done?

A) 427 B) 531
C) 651 D) 714
 
Answer & Explanation Answer: B) 531

Explanation:

Given in the question that, there are 7 boys and 6 girls. 

Team members = 5

Now, required number of ways in which a team of 5 having atleast 3 girls in the team = 

6C3  x 7C2  + 6C4 x 7C1 + 6C5= 6x5x43x2x1 x 7x62x1 + 6x5x4x34x3x2x1 x 7 + 6x5x4x3x25x4x3x2x1= 420 + 105 + 6= 531.

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5 3993
Q:

How many 3-digit numbers can be formed with the digits 1,4,7,8 and 9 if the digits are not repeated ?

A) 60 B) 26
C) 50 D) 64
 
Answer & Explanation Answer: A) 60

Explanation:

Three digit number will have unit’s, ten’s and hundred’s place.

 

Out of 5 given digits any one can take the unit’s place.

 

This can be done in 5 ways. ...              (i)

 

After filling the unit’s place, any of the four remaining digits can take the ten’s place.

 

This can be done in 4 ways. ...              (ii)

 

After filling in ten’s place, hundred’s place can be filled from any of the three remaining digits.

 

This can be done in 3 ways. ... (iii) 

 

So,by counting principle, the number of 3 digit numbers = 5x 4 x 3 = 60

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0 3947