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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:

Eight first class and six second class petty officers are on the board of the 56 club. In how many ways can the members elect, from the board, a president, a vice-president, a secretary, and a treasurer if the president and secretary must be first class petty officers and the vice-president and treasurer must be second class petty officers?

A) 1500 B) 1860
C) 1680 D) 1640
 
Answer & Explanation Answer: C) 1680

Explanation:

Since two of the eight first class petty officers are to fill two different offices, we write 8P2=56

 

Then, two of the six second class petty officers are to fill two different offices; thus, we write 6P2 =30

 

The principle of choice holds in this case; therefore, the members have 56 x 30 = 1680 ways to select the required office holders

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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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Q:

In how many ways, can zero or more letters be selected form the letters AAAAA?

A) 5 B) 6
C) 2 D) 8
 
Answer & Explanation Answer: B) 6

Explanation:

Selecting zero'A's= 1

Selecting one 'A's = 1

Selecting two 'A's = 1

Selecting three 'A's = 1

Selecting four 'A's = 1

Selecting five 'A's = 1

=> Required number ofways =6

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0 5236
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 4846
Q:

A pizza can have 3 toppings out of possible 7 toppings.How many different pizza's can be made?

A) 49 B) 35
C) 27 D) 25
 
Answer & Explanation Answer: B) 35

Explanation:

There are 7 toppings in total,and by selecting 3,we will make different types of pizza.

 

This question is a combination since having a different order of toppings will not make a different pizza.

 

So,7C3 =35

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0 5383
Q:

How many ways can you select 17 songs for mix CD out of possible 38 songs?

A) 2878 B) 2878 x 10^2
C) 290183753 D) 2878 x 10^10
 
Answer & Explanation Answer: D) 2878 x 10^10

Explanation:

since it is a combination = 38C17 = 2878 x 10^10

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0 7087
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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Q:

From a deck of 52 cards, a 7 card hand is dealt.How many distinct hands are there if the hand must contain 2 spades and 3 diamonds ?

A) 7250100 B) 7690030
C) 7250000 D) 3454290
 
Answer & Explanation Answer: A) 7250100

Explanation:

There are 13 spades,we must include 2: 13C2

 

There are 13 diamonds,we must include 3: 13C3

 

Since we can't have more than 2 spades and 3 diamonds,the remaining 2 cards must be pulled out from the 26 remaining clubs and hearts : 26C2

 

Therefore,13C2*13C3*26C2 = 7250100

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