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 How many ways can the letters of the word 'CAPITAL' be arranged in such a way that all the vowels always come together?

A) 360 B) 720
C) 120 D) 840
 
Answer & Explanation Answer: A) 360

Explanation:

CAPITAL = 7

 

Vowels = 3 (A, I, A)

 

Consonants = (C, P, T, L)

 

5 letters which can be arranged in  5P5=5!

 

Vowels A,I = 3!2!

 

No.of arrangements = 5! x 3!2!=360

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4 5187
Q:

In how many different ways  can 3 students be associated with 4 chartered accountants,assuming that each chartered accountant can take at most one student?

A) 12 B) 36
C) 24 D) 16
 
Answer & Explanation Answer: C) 24

Explanation:

Number of permutations = 4P3 = 24

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1 5177
Q:

A Cricket team of 23 people all shake hands with each other exactly once. How many hand shakes occur ?

A) 142 B) 175
C) 212 D) 253
 
Answer & Explanation Answer: D) 253

Explanation:

The first person shakes hands with 22 different people, the second person also shakes hands with 22 different people, but one of those handshakes was counted in the 22 for the first person, so the second person actually shakes hands with 21 new people. The third person, 20 people, and so on...
So,
22 + 21 + 20 + 19 + 18 + 17 + 16 + 15 + 14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= n(n+1)/2 = 22 x 23 /2 = 11 x 23 = 253.

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4 5150
Q:

In how many ways can the letters of the word 'MISSISIPPI' be arranged ?

A) 12400 B) 11160
C) 16200 D) 12600
 
Answer & Explanation Answer: D) 12600

Explanation:

Total number of alphabets = 10

so ways to arrange them = 10! 

 

Then there will be duplicates because 1st S is no different than 2nd S.

we have 4 Is 3 S and 2 Ps 

 

Hence number of arrangements = 10!/4! x 3! x 2! = 12600

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0 5114
Q:

Nine different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, how many such words can be formed which have at least one letter repeated ?

A) 43929 B) 59049
C) 15120 D) 0
 
Answer & Explanation Answer: A) 43929

Explanation:

Number of words with 5 letters from given 9 alphabets formed = 95

 

Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is = 9P5

 

Number of words can be formed which have at least one letter repeated =  95 - 9P5

 

= 59049 - 15120

 

= 43929

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11 5109
Q:

A class has 8 football players. A 5-member team and a captain will be selected out of these 8 players. How many different selections can be made ? 

A) 210 B) 168
C) 1260 D) 10!/6!
 
Answer & Explanation Answer: B) 168

Explanation:

we can select the 5 member team out of the 8 in 8C5 ways = 56 ways.

The captain can be selected from amongst the remaining 3 players in 3 ways.

Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways.

 

(or)

 

Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 28x6 = 168 ways.

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8 5048
Q:

Find the number of ways to take 20 objects and arrange them in groups of 5 at a time where order does not matter.?

A) 57090 B) 15540
C) 15504 D) 23670
 
Answer & Explanation Answer: C) 15504

Explanation:

C520 = 15504

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0 4977
Q:

Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

A) 720 B) 360
C) 120 D) 640
 
Answer & Explanation Answer: B) 360

Explanation:

Two horses A and B, in a race of 6 horses... A has to finish before B

 

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

 

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

 

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

 

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4! 

 

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

 

A cannot finish 6th, since he has to be ahead of B

 

Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

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0 4840