HackerRank Permutation and Combination Quiz -1

Two girls will always be together.

So first we select one row to seat these two girls, ways = 2. 

Now we can select two adjacent chairs in 4 ways and then girls can sit on them in 2 ways.

Ways to seat these two girls = 2 × 4 × 2 = 16 

All the girls should not be together, so at least one girl should be in another row from the remaining two girls. 

So, from the remaining two girls we select one to be seated in second row, ways = 2 

Now in the second row we select one chair out of five to seat the selected girl, ways = 5 

Ways to seat third girl = 2 × 5 = 10 

Now, the rest of the seven can be arranged in 7! Ways 

Hence, Total ways = 16 × 10 × 7! = 20 × 8! 

Total number of males = 8 and total number of females = 8.

We need to select a team of 4 persons with at least one female.

The possible combinations for a group of four persons are: 1 female and 3 males, 2 females and 2 males, 3 females and 1 male, 4 females and 0 male.

Therefore, ways in which selection can be made are = ^{8}\textrm{C}_{1}\times ^{8}\textrm{C}_{3}+ ^{8}\textrm{C}_{2}\times ^{8}\textrm{C}_{2}+ ^{8}\textrm{C}_{3}\times ^{8}\textrm{C}_{1}+ ^{8}\textrm{C}_{4}\times ^{8}\textrm{C}_{0}

= 448 + 784 + 448 + 70 = 1750.

Let the number of men be x and women be y.

In table tennis two people can play at a time.

Therefore, no of games played between men is ^{x}\textrm{C}_{2}= 36

\Rightarrow \frac{x(x-1)}{2} = 36\ \Rightarrow x(x-1)=72\ \Rightarrow x=9

 

Now, no of games played between women is ^{y}\textrm{C}_{2}= 78

\Rightarrow \frac{y(y-1)}{2} = 78\ \Rightarrow y(y-1)=156\ \Rightarrow y=13

 

Therefore, no of games in which one player is man and one is woman = ^{9}\textrm{C}_{1}\times ^{13}\textrm{C}_{1}= 117

Total number of vowels = 4

Total number of consonants = 4 

4 consonants can be arranged among themselves in 4! Ways. 

A, I, E, E and (PTNC) can be arranged in 5!/2! Ways.

∴ Number of ways = \frac{5!}{2!}\times 4! = 1440

Hence, option (A) is correct.

Number of burgers = 5, Fries = 2, cold drink = 3 and dessert = 5

∴ Possible combinations are:

Hence, Total number of possibilities = 30 × 4 × 9 × 5 = 5400

Total number of persons in an event = 20 × 5 = 100

Now every member of a family exchanged the flower with 95 other persons.

Total number of flowers exchanged = 100 × 95 = 9500

As two people exchanged two flowers with each other, the total number of flowers exchanged is half the number of flowers. 

So, total number of flowers exchanged = 9500/2 = 4750

There are 3 cousins and 8 other students. 

The number of ways to arrange these 11 students in a circular manner = (11– 1)! = 10!

These three cousins can now rearrange themselves in 3! ways. 

By the multiplication theorem, the number of ways that 3 cousins always come together in the arrangement = 8! × 3!. 

Hence, the required number of ways in which the arrangement can take place if none of the 3 cousins is seated together

= 10! – (8! × 3!) = 3628800 – (40320 * 6) 

= 3628800 – 241920 = 3386880.

The masala packets can be arranged among themselves in 6! Ways.

The Biscuit packet in 4! ways.

The Namkeen packet in 2! ways 

There are three groups which can be arranged in 3! ways. 

∴ Required number of arrangements = 6! × 4! × 2! × 3! = 720 × 24 × 2 × 6 = 207360 

= \frac{6!\ 6!\ 2}{5}

The men and women can be seated alternately so that no two women will sit together.

Hence, 5 women can be seated on alternate seats at a round table in (5 – 1)!=  4!

ways. 

Now, the 5 men can be seated in the remaining seats in 5! = 120 ways. 

Therefore the total number of ways in this case will be 5! × 4! = 2880.

There will be a director and one out of 8 faculty on either side of HoD. 

Consider the HoD and director as one unit.

The HoD and director can sit together in 2! ways.

Now, 9 people (8 faculties and 1 unit of HoD and director together) can be seated around round table in (9-1)! = 8! ways

Hence, total number of ways = 2! × 8!