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:

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 4042
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 4011
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 4004
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 3815
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 3812
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 3758
Q:

In a plane 8 points are colliner  out of 12 points, then the number of triangles we get with those 12 points is 

A) 20 B) 160
C) 164 D) 220
 
Answer & Explanation Answer: C) 164

Explanation:

For a triangle, we need 3 non-collinear points. So with 12 points (when all the 12 are such that any three non-collinear is12C3. But among them 8 points are collinear.

 

If all these 8 points are different we get 8C3 triangles as they are collinear.

 

In 12C3 triangles, we do not get 8C3 triangles

 

Therefore, The number of triangles we get = 12C3-8C3 = 164

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3 3749
Q:

Out of 6 green ball, 4 blue ball, in how many ways we selectone or more balls ?

A) 42 B) 34
C) 31 D) 22
 
Answer & Explanation Answer: B) 34

Explanation:

7 × 5 = 35

35 – 1 = 34

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6 3653