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:

$P_{r}^{n} = n (n - 1) (n - 2) . . . . (n - r + 1) = \frac{n!}{(n - r)!}$

Ex : (i) $P_{2}^{6} = (6 \times 5) = 30$ (ii) $P_{3}^{7} = (7 \times 6 \times 5) = 210$

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 $(p_{1} + p_{2} + . . . + p_{r}) = n$

Then, number of permutations of these n objects is :

$\frac{n!}{(p_{1}!) \times (p_{2}!) . . . . (p_{r}!)}$

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:

$C_{r}^{n} = \frac{n!}{(r!) (n - r)!} = \frac{n (n - 1) (n - 2) . . . . t o r f a c t o r s}{r!}$

Note : (i) $C_{n}^{n} = 1 a n d C_{0}^{n} = 1$ (ii) $C_{r}^{n} = C_{(n - r)}^{n}$

Examples : (i) $C_{4}^{11} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$ (ii) $C_{13}^{16} = C_{(16 - 13)}^{16} = C_{3}^{16} = 560$

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A) 120960	B) 120000
C) 146700	D) None of these

Answer & Explanation Answer: A) 120960

Explanation:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters = 8!/(2! x 2!)= 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =4!/2!= 12.

Required number of words = (10080 x 12) = 120960

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3 9321

A polygon 7 sides.How many diagonals can be formed?

A) 14	B) 7
C) 15	D) 21

Answer & Explanation Answer: A) 14

Explanation:

$7 C_{2}$ - 7 = 14

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3 9210

There are five cards lying on the table in one row. Five numbers from among 1 to 100 have to be written on them, one number per card, such that the difference between the numbers on any two adjacent cards is not divisible by 4. The remainder when each of the 5 numbers is divided by 4 is written down on another card (the 6th card) in order. How many sequences can be written down on the 6th card ?

A) 4 x 3^4	B) 3^4
C) 4^3	D) 3 x 4^3

Answer & Explanation Answer: A) 4 x 3^4

Explanation:

The remainder on the first card can be 0,1,2 or 3 i.e 4 possibilities.
The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier).

For each value on the card the remainder can have 3 possible values.

The total number of possible sequences is: 4*3^4

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3 9078

Find the total number of distinct vehicle numbers that can be formed using two letters followed by two numbers. Letters need to be distinct

A) 65000	B) 64000
C) 72000	D) 36000

Answer & Explanation Answer: A) 65000

Explanation:

Out of 26 alphabets two distinct letters can be chosen in $26 P_{2}$ ways. Coming to numbers part, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways.

Combined with letters there are $6 P_{2}$ X 100 ways = 65000 ways to choose vehicle numbers.

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3 9060

A box contains 4 different black balls, 3 different red balls and 5 different blue balls. In how many ways can the balls be selected if every selection must have at least 1 black ball and one red ball ?

A) $2^{4} - 1$

B) $2^{4} (2^{5} - 1)$

C) $(2^{4} - 1) (2^{3} - 1) 2^{5}$

D) None

A) A	B) B
C) C	D) D

Answer & Explanation Answer: C) C

Explanation:

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.

Hence, number of ways in which we can select the black balls

= 4C1 + 4C2 + 4C3 + 4C4
= $2^{4} - 1$ ........(A)

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls

Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
= $2^{3} - 1$ ........(B)

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.

Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + … + 5C5
= $2^{5}$ ..............(C)

From (A), (B) and (C), required number of ways
= $2^{5} (2^{4} - 1) (2^{3} - 1)$

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3 9031

Find the number of ways in which 21 balls can be distributed among 3 persons such that each person does not receive less than 5 balls.

A) 28	B) 14
C) 21	D) 7

Answer & Explanation Answer: A) 28

Explanation:

Let x, y, z be the number of balls received by the three persons, then

$x \geq 5, y \geq 5, z \geq 5 a n d x + y + z = 21$

Let $u \geq 0, v \geq 0, w \geq 0 t h e n$

$∴$ x + y + z =21

$\Rightarrow$ u + 5 + v + 5 + w + 5 = 21

$\Rightarrow$ u + v + w = 6

$∴$ Total number of solutions = ${}^{6 + 3 - 1}C_{3 - 1} = {}^{8}C_{2} = 28$

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18 8704

If the letters of the word VERMA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word VERMA is :

A) 108	B) 117
C) 810	D) 180

Answer & Explanation Answer: A) 108

Explanation:

The number of words beign with A is 4!

The number of words beign with E is 4!

The number of words beign with M is 4!

The number of words beign with R is 4!

Number of words beign with VA is 3!

Words beign with VE are VEAMR

VEARM

VEMAR

VEMRA

VERAM

VERMA

Therefore, The Rank of the word VERMA = 4 x 4! + 3! + 6 = 96 + 6 + 6 =108

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4 8454

How many number of times will the digit ‘7' be written when listing the integers from 1 to 1000?

A) 243	B) 300
C) 301	D) 290

Answer & Explanation Answer: B) 300

Explanation:

7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.

1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc

This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)

You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.

In each of these numbers, 7 is written once. Therefore, 243 times.

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77

In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).

There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.

3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.

Therefore, the total number of times the digit 7 is written between 1 and 999 is

243 + 54 + 3 = 300

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