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 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 3906
Q:

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

A) 48 B) 24
C) 64 D) 32
 
Answer & Explanation Answer: C) 64

Explanation:

Total number of balls = 2 + 3 + 4 = 9

Total number of ways 3 balls can be drawn from 9 = 9C3

No green ball is drawn = 9 - 3 = 6 = 6C3

Required number of ways if atleast one green ball is to be included = Total number of ways - No green ball is drawn

= 9C3 - 6C3

= 9x8x7/3x2  -  6x5x4/3x2

= 84 - 20

= 64 ways.

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6 3819
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 3768
Q:

Using numbers from 0 to 9 the number of 5 digit telephone numbers that can be formed is

A) 1,00,000 B) 59,049
C) 3439 D) 6561
 
Answer & Explanation Answer: C) 3439

Explanation:

The numbers 0,1,2,3,4,5,6,7,8,9 are 10 in number while preparing telephone numbers any number can be used any number of times.

 

This can be done in 105ways, but '0' is there

 

So, the numbers starting with '0' are to be excluded is 94 numbers.

 

 Total 5 digit telephone numbers = 105- 94 = 3439

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5 3754
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) 205 B) 194
C) 209 D) 159
 
Answer & Explanation Answer: C) 209

Explanation:

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


Required number of ways = (C16× C34) + C26×C24  + (C36× C14) + (C46)  

= (24+90+80+15) 

= 209.

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7 3631
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 3582
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 3576
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) 5040 B) 2525
C) 2052 D) 4521
 
Answer & Explanation Answer: A) 5040

Explanation:

The Word LOGARITHMS  contains 10 letters.

To find how many 4 letter words we can form from that = P410 =10x9x8x7 = 5040.

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3 3555