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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:

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 5291
Q:

A school committee of 5 is to be formed from 12 students.How many committees can be formed if John must be on the committee?

A) 11P4 B) 11C4
C) 11P5 D) 11C5
 
Answer & Explanation Answer: B) 11C4

Explanation:

If John must be on the committee,we have 11 students remaining,out of which we choose 4. So,11C4

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0 5276
Q:

When John arrives in New York,he has eight stops to see, but he has time only to visit six of them.In how many different ways can he arrange his schedule in New York?

A) 20610 B) 24000
C) 20160 D) 21000
 
Answer & Explanation Answer: C) 20160

Explanation:

He can arrange his schedule in 8P6 = 20160 ways

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1 5099
Q:

A standard deck of playing cards has 13 spades. How many ways can these 13 spades be arranged?

A) 13! B) 13^2
C) 13^13 D) 2!
 
Answer & Explanation Answer: A) 13!

Explanation:

The solution to this problem involves calculating a factorial. Since we want to know how 13 cards can be arranged, we need to compute the value for 13 factorial.

 

13! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800

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2 5097
Q:

There are 3 bags, in 1st there are 9 Mangoes, in 2nd 8 apples & in 3rd 6 bananas. There are how many ways you can buy one fruit if all the mangoes are identical, all the apples are identical, & also all the Bananas are identical ?

A) 23 B) 432
C) 22 D) 431
 
Answer & Explanation Answer: A) 23

Explanation:

As in this problem , buying any fruit is different case , as buying apple is independent from buying banana. so ADDITION rule will be used.

 

C19+C18+C16 = 23 will be answer.

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2 5090
Q:

Find the sum of the all the numbers formed by the digits 2,4,6 and 8 without repetition. Number may be of any of the form like 2,24,684,4862 ?

A) 133345 B) 147320
C) 13320 D) 145874
 
Answer & Explanation Answer: B) 147320

Explanation:

Sum of 4 digit numbers = (2+4+6+8) x P33 x (1111) = 20 x 6 x 1111 = 133320 


Sum of 3 digit numbers = (2+4+6+8) x P23 x (111) = 20 x 6 x 111 = 13320 


Sum of 2 digit numbers = (2+4+6+8) x P13 x (11) = 20 x 3 x 11 = 660 


Sum of 1 digit numbers = (2+4+6+8) x P03 x (1) = 20 x 1 x 1 = 20 

 

Adding All , Sum = 147320

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2 5059
Q:

From a deck of 52 cards, a 5 card hand is dealt.How may distinct five card hands are there if the queen of spades and the four of diamonds must be in the hand?

A) 52C5 B) 50C3
C) 52C4 D) 50C4
 
Answer & Explanation Answer: B) 50C3

Explanation:

If the queen of spades and the four of diamonds must be in hand,we have 50 cards remaining out of which we are choosing 3.

 

So, 50C3

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0 5029
Q:

From a deck of 52 cards, a 5 card hand is dealt.How many distinct hands can be formed if there are atleast 2 queens?

A) 103336 B) 120000
C) 108336 D) 108333
 
Answer & Explanation Answer: C) 108336

Explanation:

The total possible cases would be a 5 card hand with no restrictions :52C5 5

 

The unwanted cases are:

 

no queens(out of 48 non-queens cards we get 5) 48C5

 

only 1 queen(out of 4 queens we get 1,and out of 48 non-queens we get 4) 4C1*48C4

 

Therefore,52C5-(48C5+4C1*48C4) = 108336

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