Test: Fundamental Principle Of Counting


10 Questions MCQ Test Mathematics For JEE | Test: Fundamental Principle Of Counting


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QUESTION: 1

A lady wants to select one cotton saree and one polyster saree from a textile shop. If there are 15 cotton and 13 polyster varieties in that shop, in how many ways can she pick up two sarees?

Solution:

The lady can select one cotton saree out of 15 cotton varieties in 15 ways since
any of 15 varieties can be selected. Corresponding to each selection of a cotton saree, she can
choose a polyester saree in 13 ways. Hence the two sarees (one cotton and one polyester), by
multiplication principle of counting, can be selected in 15 x 13= 195 ways

QUESTION: 2

In a class, there are 30 boys and 18 girls. The teacher wants to select one boy and one girl to represent the class for quiz competition. In how many ways can the teacher make this selection?

Solution:

Out of 30 Boys,
No. of ways to select 1 boy = 30C1 = 30
Out of 18 Girls,
No. of ways to select 1 boy = 18C1 = 18
No. of ways of selecting 1 boy and 1 girl =
30*18 = 540

QUESTION: 3

How many numbers are there between 100 and 1000, in which all the digits are distinct?

Solution:

We know that a number between 100 and 1000 has three digits. 
So we have to find all the three digit numbers with distinct digits.

We cannot have 0 at the hundred’s place for 3-digit number.  So the hundred’s place can be filled with any of the 9 digits.  1,2,3,…..9.  So, there are 9 ways of filling the hundred’s place.

Now, 9 digits are left including 0.  So ten’s place can be filled with any of the remaining 9 digits in 9 was.  Now the unit’s place can be filled with any of the remaining 8 digits.  So, there are 8 ways of filling the unit’s place.

Hence, the total number of required numbers = 9 x 9 x 8= 648.

QUESTION: 4

How many three digit odd numbers can be formed by using the digits 1,2,3,4,5,6 if the repetition of digit is not allowed:

Solution:

There can be only 3 digits at the last digit (1,3,5) the middle digit should have 5 No's. out of 6, because one digit is used in last digit similarly 1st digit will have 4 No's.

Therefore : 4×5×3 => 60

QUESTION: 5

A class is composed 2 brothers and 6 other boys. In how many ways can all the boys be seated at the round table so that the 2 brothers are not seated besides each other?

Solution:

First we will fix one person  from  the 6 boys then 5 others can be  arranged in 5! ways = 120 ways 
Now there are 6 places left in which 2 brothers can sit,so they can choose any 2 places from the 6 places in 6C2 ways=15 ways
Also 2 brothers can arrange themselves in 2! ways
So the ways in which the two brothers can be seated =15*2  = 30
Hence total ways in which all can be seated = 120*30 = 3600

QUESTION: 6

The number of different ways in which a man can invite one or more of his 6 friends to dinner is?

Solution:

He can invite any one  friend in 6C1 ways = 6 ways:
He can invite any two friends in 6C2 ways = 15 ways
He can invite any three friends in 6C3 ways = 20 ways
He can invite any 4 friends in 6C4ways = 15 ways
He can invite any 5 friends in 6C5 ways = 6 ways
He can invite  all the 6 friends in 6C6 ways= 1 way.
Since any one of these could happen total possibilities are, 6+15+20+15+6+1 = 63.

QUESTION: 7

How many number of four digits can be formed with the digits 1,2,3,4,5 if the digit can be repeated in any number of times?

Solution:

There are 5 digits 1,2,3,4 & 5. the 4 digit number can be formed with the digits 1,2,3,4,5 where digits can be repeated =5*5*5*5=625.

QUESTION: 8

There are 7 chairs in a row. In how many ways can 3 persons occupy any three of them

Solution:

First person can sit on any of the 7 seats, second person can sit on other 6 vacant seats except one occupied by the first one. Similarly, the third can do sit on 5 so, total no. of ways = 7*6*5 = 210

QUESTION: 9

How many 4 digit numbers are there, when a digit may be repeated any number of times?

Solution:

If we assume the leading digit can not be zero, then the first 4-digit number is 1000.

The last 4-digit number is 9999.

The number of 4-digit numbers is therefore: 9999 - 1000 + 1 = 9000

QUESTION: 10

How many three digit odd numbers can be formed by using the digits 5,6,7,8 of the repetition of digits is allowed?

Solution:

The unit place can be filled in 2 ways because question asked to find 3 digit odd numbers from 5,6,7,8 and the other two places can be filled in 4 ways as repetition is allowed.
So, Unit's place can be filled in 2 ways. Decimal place can be filled in 4 ways. Hundred's place can be filled in 4 ways.
 => 4*4*2 = 32.

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