Test: Fundamental Principle Of Counting - JEE MCQ

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

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

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.

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

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

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.

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.

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

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

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.