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:

The number of signals that can be generated by using 5 differently coloured flags, when any number of them may be hoisted at a time is:

A) 235 B) 253
C) 325 D) None of these
 
Answer & Explanation Answer: C) 325

Explanation:

Required number of signals = 5P1 + 5P2 + 5P3 + 5P4 + 5P5

= 5 + 20 + 60 + 120 + 120 = 325

 

 

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4 5706
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 5635
Q:

The number of ways that 7 teachers and 6 students can sit around a table so that no two students are together is 

A) 7! x 7! B) 7! x 6!
C) 6! x 6! D) 7! x 5!
 
Answer & Explanation Answer: B) 7! x 6!

Explanation:

The students should sit in between two teachers. There are 7 gaps in between teachers when they sit in a roundtable. This can be done in 7P6ways. 7 teachers can sit in (7-1)! ways.

 

 Required no.of ways is = 7P6.6! = 7!.6!

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10 5599
Q:

In how many ways the letters of the word NUMERICAL can be arranged so that the consonants always occupy the even places ?

A) 5! B) 6!
C) 4! D) Can't determine
 
Answer & Explanation Answer: D) Can't determine

Explanation:

NUMERICAL has 9 positions in which 2, 4, 6, 8 are even positions.

And it contains 5 consonents i.e, N, M, R, C & L. Hence this cannot be done as 5 letters cannot be placed in 4 positions.

Therefore, Can't be determined.

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6 5593
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 5476
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 5472
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 5370
Q:

The Number of times the digit 8 will be written when listing the integers from 1 to 1000 is :

A) 100 B) 200
C) 300 D) 400
 
Answer & Explanation Answer: C) 300

Explanation:

 Since 8 does not occur in 1000, we have to count the number of times 8 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz, where 0x,y,z9.

 

Let us first count the numbers in which 8 occurs exactly once.

 

Since 8 can occur atone place in 3C1ways. There are 3 x 9 x 9  such numbers.

 

Next, 8 can occur in exactly two places in 3C2×9 such numbers. Lastly, 8 can occur in all three digits in one number only.

  

Hence, the number of times 8 occur is 

 1×3×92+2×3×9+3×1 = 300

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