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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:

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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83 44488
Q:

12 people at a party shake hands once with everyone else in the room.How many handshakes took place?

A) 72 B) 66
C) 76 D) 64
 
Answer & Explanation Answer: B) 66

Explanation:

There are 12 people, so this is our n value.

 

So, 12C21= 66

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89 44479
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) 4050 B) 3600
C) 1200 D) 5040
 
Answer & Explanation Answer: D) 5040

Explanation:

'LOGARITHMS' contains 10 different letters.

 

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

 

= 10P4

 

= 5040.

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43 42973
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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97 42386
Q:

In the below word how many words are there in which R and W are at the end positions?

RAINBOW

A) 120 B) 180
C) 210 D) 240
 
Answer & Explanation Answer: D) 240

Explanation:

When R and W are the first and last letters of all the words then we can arrange them in 5!ways. Similarly When W and R are the first and last letters of the words then the remaining letters can be arrange in 5! ways.

Thus the total number of permutations = 2 x 5!  = 2 x 120 = 240

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4 40000
Q:

The Indian Cricket team consists of 16 players. It includes 2 wicket keepers and 5 bowlers. In how many ways can a cricket eleven be selected if we have to select 1 wicket keeper and atleast 4 bowlers?

A) 1024 B) 1900
C) 2000 D) 1092
 
Answer & Explanation Answer: D) 1092

Explanation:

We are to choose 11 players including 1 wicket keeper and 4 bowlers  or, 1 wicket keeper and 5 bowlers.

 

Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players in 2C1*5C4*9C6 = 840

 

Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in 2C1*5C5*9C5 =252

 

Total number of ways of selecting the team = 840 + 252 = 1092

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35 34866
Q:

In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

A) 36 B) 25
C) 42 D) 120
 
Answer & Explanation Answer: A) 36

Explanation:

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

 

Let us mark these positions as under: 

                                                      (1) (2) (3) (4) (5) (6) 

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.  

Number of ways of arranging the vowels = 3P3 = 3! = 6.

 

Also, the 3 consonants can be arranged at the remaining 3 positions. 

Number of ways of these arrangements = 3P3 = 3! = 6. 

Total number of ways = (6 x 6) = 36.

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12 31096
Q:

How many 7 digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2, 4?

A) 120 B) 360
C) 240 D) 424
 
Answer & Explanation Answer: B) 360

Explanation:

There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.

 

 Number of 7 digit numbers = 7!3!×2! = 420

 

But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'.

 

=6!3!×2! = 60

 

Hence the required number of 7 digits numbers = 420 - 60 = 360

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