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:

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

Report Error

View Answer Report Error Discuss

2 4816
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

Report Error

View Answer Report Error Discuss

0 4797
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

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: CAT , Bank Exams
Job Role: Bank PO , Bank Clerk

2 4762
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.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: GRE , GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

2 4730
Q:

Determine the total number of five-card hands that can be drawn from a deck of 52 cards.

A) 2589860 B) 2598970
C) 2598960 D) 2430803
 
Answer & Explanation Answer: C) 2598960

Explanation:

When a hand of cards is dealt, the order of the cards does not matter. If you are dealt two kings, it does not matter if the two kings came with the first two cards or the last two cards. Thus cards are combinations. There are 52 cards in a deck and we want to know how many different ways we can put them in groups of five at a time when order does not matter. The combination formula is used.

C(52,5) = 2,598,960

Report Error

View Answer Report Error Discuss

1 4556
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

Report Error

View Answer Report Error Discuss

2 4512
Q:

A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

A) 1260 B) 210
C) 210 x 6! D) 1512
 
Answer & Explanation Answer: A) 1260

Explanation:

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210×6= 1260

Report Error

View Answer Report Error Discuss

0 4475
Q:

A student council of 5 members is to be formed from a selection pool of 6 boys and 8 girls.How many councils can have Jason on the council?

A) 715 B) 725
C) 419 D) 341
 
Answer & Explanation Answer: A) 715

Explanation:

If Jason is on th ecouncil,this reduces the selction pool to only 13 people,out of which we still need to select 4.

 

So, 13C4 = 715

Report Error

View Answer Report Error Discuss

1 4458