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QUESTION: 1

The number of ways in which three different rings can be worn in four fingers with at most one in each finger, are

Solution:

The total number of ways is same as the number of arrangements of 4 fingers, taken 3 at a time.

So, required number of ways = ^{4}P_{3}

= 4!/(4-3)!

= 4!/1!

= 4! => 24

QUESTION: 2

A room has 8 doors. In how many ways, a man can enter in the room through one door and exit through a different door?

Solution:

The person has 8 options to enter the hall. For each of these 8 options, he has 7 options to exit the hall. Thus, he has 8 × 7 = 56 ways to enter and exit from different doors.

QUESTION: 3

The total number of ways of answering 5 objective questions, each question having four choices are

Solution:

There are five questions

Each question has 4 options

No. of possible ways of answering each question is four

No. of Possible ways for Q1 = 4

No. of Possible ways for Q2 = 4

No. of Possible ways for Q3 = 4

No. of Possible ways for Q4 = 4

No. of Possible ways for Q5 = 4

So, Total number of ways of answering 5 objective type questions, each question having 4 choices = 4^{5}

= 1024

QUESTION: 4

In how many ways, a party of 5 men and 5 women be seated at a circular table, so that no two women are adjacent?

Solution:

Lets first place the men (M). '*' here indicates the linker of round table

* M -M - M - M - M *

which is in (5-1)! ways

So we have to place the women in between the men which is on the 5 empty seats ( 4 -'s and 1 linker i.e * )

So 5 women can sit on 5 seats in (5)! ways or

1st seat in 5 ways

2nd seat 4

3rd seat 3

4th seat 2

5th seat 1

i.e 5*4*3*2*1 ways

So the answer is 5! * 4! = 2880

QUESTION: 5

The measure of an interior angle of a regular polygon is 140°. The number of sides and diagonals in this polygon are:

Solution:

Since the exterior angle 140 degrees, The sum of the interior angles = (2n - 4)* right angles. So 140n = (2n - 4)* right angles, or

140n = (2n - 4)*90, or

140n = 180n - 360o, or

40n = 360°, or

n = 9 sides.

QUESTION: 6

In how many ways can a cricket team of 11 players be chosen out from a squad of 14 players, if 5 particular players are always chosen?

Solution:

Total no of players = 14 out of which 5 are fixed.

So, 11-5 = 6

Remaining players = 14 - 6

= 9 players

9C6 = 9!/(3!*6!)

= 84

QUESTION: 7

A team of 7 players is to be formed out of 5 under 19 players and 6 senior players. In how many ways, the team can be chosen when at least 4 senior players are included?

Solution:

No. of ways to select 4 senior and 3 U-19 players = ^{6}C_{4} * ^{5}C3 = 150

No. of ways to select 5 senior and 2 U-19 players = ^{6}C_{5} * ^{5}C2 = 60

No. of ways to select 6 senior and 1 U-19 players = ^{6}C_{6} * ^{5}C1 = 5

Total no. of ways to select the team = 150 + 60 + 5 = 215

QUESTION: 8

Four alphabets A, M, P, O are purchased from a warehouse. How many ordered pairs of initials can be formed using these?

Solution:

Total number of letters = 4

Number of ordered pairs of letters that can be formed like (A, M) or (P, O) etc = ^{4}P_{2} = 4!/2!

= 24/2

= 12

QUESTION: 9

How many words beginning with ‘T’ and ending with ‘E’ can be formed using the letters of the word”TRIANGLE” ?

Solution:

There are 8 letters in the word TRIANGLE.

2 alphabets are fixed, remaining are 6 alphabets

So, number of arrangements = ^{6}P_{6}

= 6!

= 720

QUESTION: 10

In how many ways can 4 red, 3 yellow and 2 green chairs be arranged in a row if the chairs of the same colour are indistinguishable?

Solution:

Total no of balls = 9

red balls = 4

yellow balls = 3

green balls = 2

Total no. of arrangements = 9!/(4!*3!*2!)

= 1260

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