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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:

How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can take more than 10 letters ?

A) 5^10 B) 10^5
C) 5P5 D) 5C5
 
Answer & Explanation Answer: A) 5^10

Explanation:

Each of the 10 letters can be posted in any of the 5 boxes.

 

So, the first letter has 5 options, so does the second letter and so on and so forth for all of the 10 letters.

 

i.e. 5*5*5*….*5 (upto 10 times) = 5 ^ 10.

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14 22989
Q:

How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane?

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

Explanation:

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

 

AB , AC

 

BA , BC

 

CA , CB

 

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.

 

The lines are: AB, BC and AC ; 3 lines only.

 

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

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Q:

In how many ways can the letters of the word "PROBLEM" be rearranged to make 7 letter words such that none of the letters repeat?

A) 49 B) 7!
C) 7^7 D) 7^3
 
Answer & Explanation Answer: B) 7!

Explanation:

There are seven positions to be filled.

 

The first position can be filled using any of the 7 letters contained in PROBLEM.

 

The second position can be filled by the remaining 6 letters as the letters should not repeat.

 

The third position can be filled by the remaining 5 letters only and so on.

 

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.

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2 14018
Q:

In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?

A) 4!/2! B) 3!/2!
C) (4! x 3!) / 2! D) 36
 
Answer & Explanation Answer: C) (4! x 3!) / 2!

Explanation:

ABACUS is a 6 letter word with 3 of the letters being vowels.

 

If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together.

 

These 4 elements can be rearranged in 4! Ways.

 

The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a" appears twice.

 

Hence, the total number of rearrangements in which the vowels appear together are (4! x 3!)/2!

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3 10208
Q:

How many different four letter words can be formed (the words need not be meaningful using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R?

A) 59 B) 56
C) 64 D) 55
 
Answer & Explanation Answer: A) 59

Explanation:

The first letter is E and the last one is R.

 

Therefore, one has to find two more letters from the remaining 11 letters.

 

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

 

The second and third positions can either have two different letters or have both the letters to be the same.

 

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

 

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

 

Total number of possibilities = 56 + 3 = 59

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22 20945
Q:

A certain marathon had 50 people running for first prize, second, and third prize.How many ways are there to correctly guess the first, second, and third place winners?

A) 2 B) 1
C) 4 D) 3
 
Answer & Explanation Answer: B) 1

Explanation:

There is 1 way to correctly guess who comes in first, second, and third. There is only one set of first, second and third place winners. You must correctly guess these three people, and there is only one way to do so.

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0 6036
Q:

Find the number of ways to draw a straight, (suit does not matter) beginning with a 4 and ending with a 8?

A) 1024 B) 1296
C) 1094 D) 1200
 
Answer & Explanation Answer: A) 1024

Explanation:

There are 5 slots.

 

                   __ __ __ __ __

 

The first slot must be a four. There are 4 ways to put a four in the first slot.

 

There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc.

 

(4)(4)(4)(4)(4) = 1024

 

Therefore there are 1024 different ways to produce the desired hand of cards.

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0 4159
Q:

How many ways are there to deal a five-card hand consisting of three eight's and two sevens?

A) 36 B) 72
C) 24 D) 16
 
Answer & Explanation Answer: C) 24

Explanation:

If a card hand that consists of four Queens and an Ace is rearranged, nothing has changed.

 

The hand still contains four Queens and an Ace. Thus, use the combination formula for problems with cards.

 

We have 4 eights and 4 sevens.

We want 3 eights and 2 sevens.

C(have 4 eights, want 3 eights) x C(have 4 sevens, want 2 sevens) 

C(4,3) x C(4,2) = 24

 

Therefore there are 24 different ways in which to deal the desired hand.

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