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 Permutations and Combinations Questions

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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:

Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements.

A) 340 B) 370
C) 320 D) 330
 
Answer & Explanation Answer: D) 330

Explanation:

Here the order of choosing the elements doesn’t matter and this is a problem in combinations.

 

We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

 

This can be done in 11C4 ways = 330 ways

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15 16122
Q:

The number of ways that 8 beads of different colours be strung as a necklace is 

A) 2520 B) 2880
C) 4320 D) 5040
 
Answer & Explanation Answer: A) 2520

Explanation:

The number of ways of arranging n beads in a necklace is (n-1)!2=(8-1)!2=7!2 = 2520 

(since n = 8)

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27 15785
Q:

A coach must choose five starters from a team of 12 players. How many different ways can the coach choose the starters ?

A) 569 B) 729
C) 625 D) 769
 
Answer & Explanation Answer: B) 729

Explanation:

Choose 5 starters from a team of 12 players. Order is not important.

 

12C5= 729

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7 15767
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 15261
Q:

How many 6-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that the digits should not repeat and the second last digit is even ?

A) 521 B) 720
C) 420 D) 225
 
Answer & Explanation Answer: B) 720

Explanation:

Let last digit is 2
when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.
so for last digit = 2, total numbers=240

 

Similarly for 4 and 6
When last digit = 4, total no. of ways =240
and last digit = 6, total no. of ways =240
so total of 720 even numbers are possible.

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13 14998
Q:

How many different words can be made using the letters of the word ' HALLUCINATION ' if all constants are together?

A) 129780 B) 1587600
C) 35600 D) None of these
 
Answer & Explanation Answer: B) 1587600

Explanation:

 H   L   C   N    T    A   U   I   O

 

      L        N          A        I

 

There are total 131 letters out of which 7 are consonants and 6 are vowels. Also ther are 2L's , 2N's, 2A's and 2I's.

 

If all the consonants  are together then the numberof arrangements = 7!2!x 1/2! .

But the 7 consonants  can be arranged themselves in  7!2! x 1/2! ways. 

 

Hence the required number of ways = 7!2!2!2 = 1587600

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7 14964
Q:

Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?

A) 2400 B) 6400
C) 3600 D) 7200
 
Answer & Explanation Answer: C) 3600

Explanation:

Let the beds be numbered 1 to 7.

 

Case 1 : Suppose Anju is allotted bed number 1. 

Then, Parvin cannot be allotted bed number 2. 

So Parvin can be allotted a bed in 5 ways. 

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5´5!ways = 600 ways.

 

Case 2 : Anju is allotted bed number 7. 

Then, Parvin cannot be allotted bed number 6 

As in Case 1, the beds can be allotted in 600 ways.

 

Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6. 

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways. 

The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways

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15 14916
Q:

12 points lie on a circle. How many cyclic quadrilaterals can be drawn by using these points?

A) 500 B) 490
C) 495 D) 540
 
Answer & Explanation Answer: C) 495

Explanation:

For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4 points out of 12 points is 12C4 = 495.

 

Therefore, we can draw 495 quadrilaterals

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3 14874