Permutation and combination questions

Permutation And Combination

Question 1
A book-shelf can accommodate 6 books from left to right. If 10 identical books on each of the languages A,B,C and D are available, In how many ways can the book shelf be filled such that book on the same languages are not put adjacently.
A
(6P4) /2!
B
$4\times 3^5$
C
(40P6) /6!
D
None of the above
Question 1 Explanation: 
First place can be filled in 4 ways.

The subsequent places can be filled in 3 ways each.

Hence, the number of ways =4 *3 *3 *3 *3 *3 = 4 *35







Question 2
How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
A
5040
B
40
C
2520
D
400
Question 2 Explanation: 
'LOGARITHMS' contains 10 different letters.

Required number of words

= Number of arrangements of 10 letters, taking 4 at a time.

= 10P4

= (10 x 9 x 8 x 7)

= 5040.

Question 3
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
A
15
B
5
C
10
D
20
Question 3 Explanation: 
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

∴ Required number of numbers = (1 x 5 x 4) = 20.







Question 4
How many positive integers 'n' can be form using the digits 3,4,4,5,6,6,7, if we want 'n' to exceed 60,00,000?
A
360
B
320
C
720
D
540
Question 4 Explanation: 
As per the given condition, number in the highest position should be either 6 or 7, which can be done in 2 ways.

If the first digit is 6, the other digits can be arranged in 6! /2! = 360 ways.

If the first digit is 7, the other digits can be arranged in 6!/(2!$\times$2!)=180 ways.

Thus required possibilities for n,

= 360 +180 = 540 ways.

Question 5
In a hockey championship, there are 153 matches played. Every two team played one match with each other. The number of teams participating in the championship is:
A
19
B
18
C
16
D
17
Question 5 Explanation: 
Let there were x teams participating in the games, then total number of matches,

nC2 = 153.

On solving we get,

=> n =-17 and n =18.

It cannot be negative so,

n = 18 is the answer.







Question 6
There are 6 equally spaced points A,B,C,D,E and F marked on a circle with radius R. How many convex pentagons of distinctly different areas can be drawn using these points as vertices?
A
None of these
B
1
C
6P5
D
5
Question 6 Explanation: 
Since, all the points are equally spaced; hence the area of all the convex pentagons will be same.
Question 7
12 chairs are arranged in a row and are numbered 1 to 12. 4 men have to be seated in these chairs so that the chairs numbered 1 to 8 should be occupied and no two men occupy adjacent chairs. Find the number of ways the task can be done.
A
432
B
360
C
384
D
470
Question 7 Explanation: 
Given there are 12 numbered chairs, such that chairs numbered 1 to 8 should be occupied.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

The various combinations of chairs that ensure that no two men are sitting together are listed.

(1, 3, 5,...), The fourth chair can be 5,6,10,11 or 12, hence 5 ways.

(1, 4, 8, ...), The fourth chair can be 6,10,11 or 12 hence 4 ways.

(1, 5, 8, ...), the fourth chair can be 10,11 or 12 hence 3 ways.

(1, 6, 8,...), the fourth chair can be 10,11 or 12 hence 3 ways.

(1,8,10,12) is also one of the combinations.

Hence, 16 such combinations exist.

In case of each these combinations we can make the four men inter arrange in 4! ways.

Hence, the required result =16$\times$4!= 384.







Question 8
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
A
480
B
720
C
360
D
5040
Question 8 Explanation: 
The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Therefore Required number of ways = (120 x 6) = 720.

Question 9
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
210 *33
B
42 *33
C
4 *34
D
210
Question 9 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 vales (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 *34.







Question 10
How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different?
A
12C3 *4C4
B
16C7 *7!
C
12C4 *4C3 *7!
D
11C4 *4C3
Question 10 Explanation: 
4 consonants out of 12 can be selected in,

12C4 ways.

3 vowels can be selected in 4C3 ways.

Therefore, total number of groups each containing 4 consonants and 3 vowels,

= 12C4 *4C3 Each group contains 7 letters, which can be arranging in 7! ways.

Therefore required number of words,

= 12C4 *4C3 *7!

There are 10 questions to complete.

Leave a Reply

Your email address will not be published. Required fields are marked *