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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:

A research team of 6 people is to be formed from 10 chemists,5 politicians, 8 economists and 15 biologists.How many teams have atleast 5 chemists?

A) 7350 B) 6400
C) 6379 D) 7266
 
Answer & Explanation Answer: D) 7266

Explanation:

10C5*28C1*10C6 = 7266

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1 6719
Q:

In how many ways the letters of the word OLIVER be arranged so that the vowels in the word always occur in the dictionary order as we move from left to right ?

A) 186 B) 144
C) 136 D) 120
 
Answer & Explanation Answer: D) 120

Explanation:

In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O.

 

There are 6 letters in thegiven word.

 

First arrange 3 vowels.

 

This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O)

 

Remaining 3 letters can be placed in 3 places = 3! ways

 

Total number of possible ways of arranging letters of OLIVER = 3! x C36 ways = 6x5x4 = 120 ways.

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11 6702
Q:

Jay wants to buy a total of 100 plants using exactly a sum of Rs 1000. He can buy Rose plants at Rs 20 per plant or marigold or Sun flower plants at Rs 5 and Re 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase?

A) 3 B) 6
C) 4 D) 2
 
Answer & Explanation Answer: A) 3

Explanation:

Let the number of Rose plants be ‘a’.
Let number of marigold plants be ‘b’.
Let the number of Sunflower plants be ‘c’.
20a+5b+1c=1000; a+b+c=100

 

Solving the above two equations by eliminating c,
19a+4b=900

b = (900-19a)/4 

b = 225 - 19a/4----------(1)


b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2)

Substituting (1) in (2),

 0 < 225 - 19a/4 < 99

225 <  -19a/4 < (99 -225)

=> 4 x 225 > 19a > 126 x 4

=> 900/19 > a > 505

 

a is the integer between 47 and 27 ----------(3)
From (1), it is clear, a should be multiple of 4.


Hence possible values of a are (28,32,36,40,44)


For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40


Three solutions are possible.

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2 6643
Q:

There are fourteen juniors and twenty-three seniors in the Service Club. The club is to send four representatives to the State Conference. If the members of the club decide to send two juniors and two seniors, how many different groupings are possible ?

A) 23024 B) 24023
C) 23023 D) 25690
 
Answer & Explanation Answer: C) 23023

Explanation:

Choose 2 juniors and 2 seniors.

 

14C2*23C2 = 23023

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1 6599
Q:

In how many different ways can five persons stand in a line for a group photograph?

A) 120 B) 240
C) 360 D) 720
 
Answer & Explanation Answer: A) 120

Explanation:

This is the number of permutations of five things taken all at a time.

 

Therefore, answer = 5P5 = 120 ways

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2 6571
Q:

When six fairs coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads ?

A) 15 B) 42
C) 16 D) 40
 
Answer & Explanation Answer: B) 42

Explanation:

The question requires you to find number of the outcomes in which at most 3 coins turn up as heads.

i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.

 

The number of outcomes in which 0 coins turn heads is 6C0=1 

The number of outcomes in which 1 coin turns head is =6C1=6 

The number of outcomes in which 2 coins turn heads is6C2=15 

The number of outcomes in which 3 coins turn heads is6C3=20

 

Therefore, total number of outcomes =1+6+15+20= 42 outcomes

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3 6560
Q:

Find the number of ways to arrange 6 items in groups of 4 at a time where order matters?

A) 720 B) 640
C) 740 D) 360
 
Answer & Explanation Answer: D) 360

Explanation:

6P4 = 6! / (6-4)! = 360

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0 6435
Q:

In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?

A) 15 B) 10
C) 25 D) 20
 
Answer & Explanation Answer: C) 25

Explanation:

The toys are different; The boxes are identical
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways
a. 2,2,1
b. 3,1,1

Case a. Number of ways of achieving the first option 2−2–1

Two toys out of the 5 can be selected in5C2 5 ways. Another 2 out of the remaining 3 can be selected in3C2 ways and the last toy can be selected in 1C1  way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2.

Therefore, total number of ways of achieving the 2−2–1 option is:

5C2*3C22=10*32=15 ways.

Case b. Number of ways of achieving the second option 3−1–1

Three toys out of the 5 can be selected in 5C3  ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3−1–1 option is5C3=10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes
=number of ways of achieving Case a + number of ways of achieving Case b
=15 + 10 = 25 ways

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0 6390