Quants Questions for CLAT | QB Set 23

A college is organising an Inter-College Leadership Summit and has formed a student committee to manage the event. The committee has 8 students, including Aarav, Bhavna, Charu, Dev, Esha, Farhan, Gauri and Harsh. Different arrangements and selections must be made for the summit.

For the opening ceremony, the organisers have to select 3 students out of the 8 to welcome the guests. Since all three students will perform the same duty, the order in which they are selected does not matter. The organisers also have to choose a President, Vice-President and Secretary from among the 8 students. As each position is different, the order of selection becomes important.

For another activity, 5 students are required to sit in a row on the stage for a group presentation. The organisers are also planning a quiz team consisting of 4 students from the same group of 8. In addition, four specifically selected students—Aarav, Bhavna, Charu and Dev—may be asked to stand in a straight line for photographs.

Using the principles of permutation and combination, answer the following questions.

Questions

Question 1

In how many different ways can 3 students be selected from the 8 students to welcome the guests?

A. 24

B. 48

C. 56

D. 336

Question 2

In how many ways can the positions of President, Vice-President and Secretary be assigned among the 8 students?

A. 56

B. 168

C. 336

D. 512

Question 3

In how many different ways can 5 students be selected and arranged in a row from the group of 8 students?

A. 672

B. 3,360

C. 6,720

D. 40,320

Question 4

How many different quiz teams of 4 students can be formed from the group of 8?

A. 70

B. 56

C. 168

D. 280

Question 5

Aarav, Bhavna, Charu and Dev have been selected for a photograph. In how many ways can they stand in a row if Aarav and Bhavna must always stand together?

A. 6

B. 8

C. 16

D. 12

Answers and Detailed Explanations

Question 1: Answer – C. 56

The organisers need to select 3 students from 8, and all three students perform the same role. Therefore, the order of selection does not matter.

Hence, combination is used:

8C3 = 8! / (3! × 5!)

= (8 × 7 × 6) / (3 × 2 × 1)

= 336 / 6 = 56

Therefore, there are 56 different ways to select the three students.

Question 2: Answer – C. 336

Here, three different positions have to be assigned: President, Vice-President and Secretary.

Since the positions are different, the order matters. Aarav as President and Bhavna as Vice-President is different from Bhavna as President and Aarav as Vice-President.

Therefore, permutation is used:

8P3 = 8! / 5!

= 8 × 7 × 6 = 336

Thus, the three positions can be assigned in 336 ways.

Question 3: Answer – C. 6,720

Five students have to be selected from 8 and arranged in a row.

Since their positions in the row matter, permutation must be used.

8P5 = 8! / 3!

= 8 × 7 × 6 × 5 × 4

= 6,720

Another way to understand this is that there are 8 choices for the first position, 7 for the second, 6 for the third, 5 for the fourth and 4 for the fifth. Therefore, 8 × 7 × 6 × 5 × 4 = 6,720.

Question 4: Answer – A. 70

The organisers have to form a quiz team of 4 students from 8 students.

There are no separate positions within the team. Therefore, selecting Aarav, Bhavna, Charu and Dev represents the same team regardless of the order in which their names are chosen.

Hence, combination is used:

8C4 = 8! / (4! × 4!)

= (8 × 7 × 6 × 5) / (4 × 3 × 2 × 1)

= 1,680 / 24 = 70

Therefore, 70 different quiz teams can be formed.

Question 5: Answer – D. 12

Aarav, Bhavna, Charu and Dev have to stand in a row, but Aarav and Bhavna must remain together.

Treat Aarav and Bhavna as a single unit. The units are now: the Aarav-Bhavna block, Charu and Dev.

These 3 units can be arranged in 3! = 6 ways.

Within the Aarav-Bhavna block, the two students can stand as Aarav-Bhavna or Bhavna-Aarav. Thus, the block has 2! = 2 internal arrangements.

Total arrangements = 3! × 2! = 6 × 2 = 12.

Thus, they can stand in 12 different ways while Aarav and Bhavna remain together.