In a row, there are 81 rooms, whose door no. are 1 , 2 , … … . , 81 , initially all the door are closed. A person takes 81 round of the row, numbers as 1 st round, 2 nd round … … . . 81 th round. In each round, he interchage the position of those door number, whose number is multiple of the round number. Find out after 81 st round, How many doors will be open.

In the question, a door will toggle whenever it is multiple of the given round for example in 1st round, every door will toggle because every number is multiple of 1 and in 2 nd round 2 , 4 , 6 . . . , only even door will toggle 3 = 1 , 3 4 = 1 , 2 , 4 So, here above we see that any door will toggle upto its number of divisors, so if the number of divisors is even, door will be closed in the end and if the number of divisors are odd, door will be open So the answer is all perfect squares from 1 to 81 . 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81

In a row, there are 81 rooms, whose door no. are 1 , 2 , … … . , 81 , initially all the door are closed. A person takes 81 round of the row, numbers as 1 st round, 2 nd round … … . . 81 th round. In each round, he interchage the position of those door number, whose number is multiple of the round number. Find out after 81 st round, How many doors will be open.

In the question, a door will toggle whenever it is multiple of the given round for example in 1st round, every door will toggle because every number is multiple of 1 and in 2 nd round 2 , 4 , 6 . . . , only even door will toggle 3 = 1 , 3 4 = 1 , 2 , 4 So, here above we see that any door will toggle upto its number of divisors, so if the number of divisors is even, door will be closed in the end and if the number of divisors are odd, door will be open So the answer is all perfect squares from 1 to 81 . 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81

In how many ways can two distinct subsets of the set A of k k ≥ 2 elements be selected so that they have exactly two common elements.

Total elements = k so, every element has 3 choices . Choice 1 is to go in the first subset, choice 2 is to go in the second subset and choice 3 is to go neither of them. but, according to question, subset 1 and subset 2 both have 2 elements in common. Choosing 2 elements in common by C 2 k ways. The ways by which elements will go in different subsets = 3 k - 2 - 1 (we subtract 1 because of the case when both of the sets are empty) Now, every case is repeating twice so the final answer will be half of the total number of ways ⇒ C 2 × 3 k - 2 - 1 k 2 ⇒ k ! 2 ! k - 2 ! × 3 k - 2 - 1 2 ⇒ k k - 1 4 × 3 k - 2 - 1

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 8 - Permutation and Combination>Exercise-4>Q 16

MEDIUM

JEE Main/Advance

IMPORTANT

Earn 100

In a row, there are $81$ rooms, whose door no. are $1, 2, \dots \dots ., 81,$ initially all the door are closed. A person takes $81$ round of the row, numbers as $1^{st}$ round, $2^{nd}$ round $\dots \dots . . 81^{th}$ round. In each round, he interchage the position of those door number, whose number is multiple of the round number. Find out after $81^{st}$ round, How many doors will be open.

Important Questions on Permutation and Combination

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 8 - Permutation and Combination>Exercise-4>Q 17

Mr. Sibbal walk up

16

steps, going up either

1

or

2

steps with each stride there is explosive material on the

8^{th}

step so he cannot step there. Then number of ways in which Mr. Sibbal can go up.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 8 - Permutation and Combination>Exercise-4>Q 18

A batsman scores exactly a century by hitting fours and sixes in twenty consecutive balls. In how many different ways can he hit either six or four or play a dot ball