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Test: Permutation And Combinations (Competition Level) - 1


30 Questions MCQ Test Mathematics (Maths) Class 11 | Test: Permutation And Combinations (Competition Level) - 1


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This mock test of Test: Permutation And Combinations (Competition Level) - 1 for JEE helps you for every JEE entrance exam. This contains 30 Multiple Choice Questions for JEE Test: Permutation And Combinations (Competition Level) - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Permutation And Combinations (Competition Level) - 1 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Permutation And Combinations (Competition Level) - 1 exercise for a better result in the exam. You can find other Test: Permutation And Combinations (Competition Level) - 1 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Solution:

Selecting 3 consonants out of 7,
No. of ways = 7C3 
Selecting 2 vowels out of 4,
No. of ways = 4C2 
No. of ways of forming the words
= 7C3* 4C2* 5! (for arrangement of the letters)
= 35 * 6 * 120
= 25200 

QUESTION: 2

 Four visitors A, B, C & D arrive at a town which has 5 hotels. In how many ways can they disperse themselves among 5 hotels, if 4 hotels are used to accommodate them.


Solution:

Given, 
Number of visitors= 4
Number of hotels= 5
So, A can disperse among 5 hotels,
      B can disperse among 4 hotels,
      C can disperse among 3 hotels,
      D can disperse among 2 hotels.
∴ They can disperse in the hotels in 5 × 4 × 3 × 2 = 120 ways

QUESTION: 3

If the letters of the word "VARUN" are written in all possible ways and then are arranged as in a dictionary, then rank of the word VARUN is

Solution:

Number of words beginning with A = 4! = 4 × 3 × 2 = 24
Number of words beginning with N = 4! = 4 × 3 × 2 = 24
Number of words beginning with R = 4! = 4 × 3 × 2 = 24
Number of words beginning with U = 4! = 4 × 3 × 2 = 24
So, in total 96 words will be formed while beginning with letter A, N, R and U.
Now, 97th word according to dictionary order will be VANRU
Now, 98th word according to dictionary order will be VANUR
Now, 99th word according to dictionary order will be VARNU
Finally, 100th word according to dictionary order will be VARUN.
Hence, option (c) is the correct answer.

QUESTION: 4

How many natural numbers are their from 1 to 1000 which have none of their digits repeated.

Solution:
QUESTION: 5

 A man has 3 jackets, 10 shirts, and 5 pairs of slacks. If an outfit consists of a jacket, a shirt, and a pair of slacks, how many different outfits can the man make ?

Solution:

Total number of jackets=3
Total number of shirts=10
Total number of pair of slacks=5
Ways of selecting one jacket =3C1=3
Ways of selecting one shirt =10C1=10
Ways of selecting one pair of slack = 5C1=5
Number of different outfits that can be made=3×10×5=150
Number of different outfits that can be made=150

QUESTION: 6

There are 6 roads between A & B and 4 roads between B & C.
(i) In how many ways can one drive from A to C by way of B?
(ii) In how many ways can one drive from A to C and back to A, passing through B on both trips?
(iii) In how many ways can one drive the circular trip described in (ii) without using the same road more than once.
Solution:

i) It's 6*4 because there are 6 different roads to get to B, then after getting there, can take 4 different paths. So every one road to B has 4 different possible paths. 

 

 

QUESTION: 7

How many car number plates can be made if

i) each plate contains 2 different letters of English alphabet, followed by 3 different digits.

ii) the number plates have two letters of the english alphabet followed by two digits if the repetition of digits or alphabets is not allowed?
Solution:

i) There are 5 places to fill

2 places of alphabet(out of 26) can be filled in =26∗25 ways

3 places of digits(out of 10) can be filled in = 10∗9∗8 ways

No of different plates = 26∗25∗10∗9∗8=468000

 

ii) We have total four place to cover with two alphabets and two digits.

For the first place:

Out of 26 alphabets we need one  

∴ combinations for first place = 26C1 = 26

For the second place:

We are left with 25 alphabets now as no repetitions are allowed.

∴combinations for second place= 25C1 = 25

For the third place:

Out of 10 digits we need to select one

∴combinations for third place = 10C1 = 10

For the fourth place:

We are left with 9 digits now as no repetitions are allowed

∴combinations for fourth place= 9C1 = 9

∴Total Combinations of number plates = 26 * 25 * 10 * 9 = 58,500

QUESTION: 8

(i) Find the number of four letter word that can be formed from the letters of the word HISTORY. (each letter to be used at most once) 

(ii) How many of them contain only consonants? 

(iii) How many of them begin & end in a consonant? 

(iv) How many of them begin with a vowel? 

(v) How many contain the letters Y? 

(vi) How many begin with T & end in a vowel? 

(vii) How many begin with T & also contain S? 

(viii) How many contain both vowels?
Solution:

QUESTION: 9

If repetitions are not permitted

(i) How many 3 digit numbers can be formed from the six digits 2, 3, 5, 6, 7 & 9 ?

(ii) How many of these are less than 400 ?

(iii) How many are even ?

(iv) How many are odd ?
Solution:

i) 6!/(6-3)!=6*5*4=120

.

ii) 2/6 * 120 = 40

.

iii) 2/6 * 120 = 40

.

iv) 1/6 * 120 = 20

QUESTION: 10

 In how many ways can 5 letters be mailed if there are 3 different mailboxes available if each letter can be mailed in any mailbox.

Solution:

Different number of ways = 35
= 243

QUESTION: 11

 Every telephone number consists of 7 digits. How many telephone numbers are there which do not include any other digits but 2, 3, 5, & 7 ?

Solution:
QUESTION: 12

In a telephone system four different letter P, R, S, T and the four digits 3, 5, 7, 8 are used.Find the maximum number of “telephone numbers” the system can have if each consistsof a letter followed by a four-digit number in which the digit may be repeated.

Solution:
QUESTION: 13

 How many of the arrangements of the letter of the word "LOGARITHM" begin with a vowel and end with a consonant ?

Solution:
QUESTION: 14

Number of natural numbers between 100 and 1000 such that at least one of their digits is 7, is
Solution:

A 3-digit no.: _ _ _
Possible digits: 0, 1 ,2, 3, 4, 5, 6, 7, 8, 9
In the first place there are 9 possible digits (0 cannot be the 1st digit)
In the second and third place there are 10 possible digits.
Total no. of 3-digit numbers = 9 *10 *10 = 900 
For 3-digit no. without digit 7,
Possible digits: 0, 1 ,2, 3, 4, 5, 6, 8, 9
In the first place there are 8 possible digits (0 cannot be the 1st digit)
In the second and third place there are 9 possible digits.
Total no. of 3-digit numbers = 8 *9 *9 = 648 
No. of 3-digit numbers with at least one digit 7 = 900 – 648 = 252

QUESTION: 15

How many four digit numbers are there which are divisible by 2.

Solution:
QUESTION: 16

 In a telephone system four different letter P, R, S, T and the four digits 3, 5, 7, 8 are used. Find the maximum number of "telephone numbers" the system can have if each consists of a letter followed by a four-digit number in which the digit may be repeated.

Solution:
QUESTION: 17

Find the number of 5 lettered palindromes which can be formed using the letters from the English alphabets.

Solution:

For 5 lettered palindrome, we can fill the first 3 letters randomly and then the last 2 letters will become fixed.

there are 26 letters.
therefore the number of ways =26×26×26=263
 = 17,576.

QUESTION: 18

 Number of ways in which 7 different colours in a rainbow can be arranged if green is always in the middle.

Solution:
QUESTION: 19

Two cards are drawn one at a time & without replacement from a pack of 52 cards. Determine the number of ways in which the two cards can be drawn in a definite order.

Solution:

We are drawing two cards without replacement from a deck of 52 cards.

No. of ways = 52C1 x 51C1 = 52 x 51 = 2652

QUESTION: 20

Numbers of words which can be formed using all the letters of the word "AKSHI", if each word begins with vowel or terminates in vowel.

Solution:

Consider the words starting with A.
A−−−−
Number of ways of filling the 2nd place is 4
Number of ways of filling the 3rd place is 3
Number of ways of filling the 4th place is 2
Number of ways of filling the 5th place is 1
Hence in total 4×3×2×1=24
Similar cases for the words starting with I.=24
Hence 2×24 =48
Consider the words ending with A−−−−A
Number of ways of filling the 1st place is 4
Number of ways of filling the 2nd place is 3
Number of ways of filling the 3rd place is 2
Number of ways of filling the 4th place is 1
Hence in total 24 ways.
Similarly for I- 24 ways.
Hence total number of words 4×24=96
But this includes the words which start and end with Vowels.
Hence A−−−I
Number of such words is 3!.
Similarly I−−−A.
Number of such words is 3!.
Hence total number of words beginning with or ending with Vowels is
= 96−3!−3!=96−12
= 84

QUESTION: 21

 A letter lock consists of three rings each marked with 10 different letters. Find the number of ways in which it is possible to make an unsuccessful attempts to open the lock.

Solution:
QUESTION: 22

 How many 10 digit numbers can be made with odd digits so that no two consecutive digits are same.

Solution:

The odd digits which can be considered are (1,3,5,7 and 9).
If we start from left, the first digit can be any number out of the five digits. The second place can be filled with any four digits except the digit at the first place.
Similarly, four ways for the third digit until we reach the tenth place.
Therefore, the total numbers formed are 5 × 49

QUESTION: 23

 It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

Solution:
QUESTION: 24

If no two books are alike, in how many ways can 2 red, 3 green, and 4 blue books be arranged on a shelf so that all the books of the same colour are together ?

Solution:
QUESTION: 25

In how many ways can 5 girls be seated in a row of 5 seats?
Solution:

There can 5 girls selected for the first seat.

Then, there are 4 girls to select from for the second seat.

Then, there are 3 girls to select from for the third seat.

Then, there are just 2 girls to select from for the fourth seat.

Then, there is just 1 girl for the fifth and final seat.

Therefore they number of ways to seat 5 girls in 5 seats is:

5×4×3×2×1⇒20×6×1⇒120×1⇒120

So there are 120 different ways to seat 5 girls in 5 chairs.

QUESTION: 26

How many of the 900 three digit numbers have at least one even digit ?
Solution:

Total 3 digit numbers = 900
There are only 5 odd numbers (1,3,5,7,9)
So 3 digit numbers having containing only odd digits = 5*5*5 = 125
So, 3 digit numbers containing at least one even digit = 900–125 = 775

QUESTION: 27

The number of natural numbers from 1000 to 9999 (both inclusive) that do not have all 4 different digits is
Solution:
QUESTION: 28

The number of different seven digit numbers that can be written using only three digits 1, 2 & 3 under the condition that the digit 2 occurs exactly twice in each number is
Solution:
QUESTION: 29

Out of seven consonants and four vowels, the number of words of six letters, formed by taking four consonants and two vowels is (Assume that each ordered group of letter is a word)
Solution:
QUESTION: 30

All possible three digits even numbers which can be formed with the condition that if 5 is one of the digit, then 7 is the next digit is :
Solution:

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