16+ Factorial Questions for CAT with SOLUTIONS

Tabulation Questions for the CAT are part of the Quantitative Aptitude Through Tabulation questions, aspirants are tested to solve questions with systematic demonstration. The difficulty level of these questions is moderate to difficult.

Study the given table carefully and answer the questions that follow :

Question 1: A software company needs to form a team of 6 developers for a project. However, there are 10 developers available. In how many ways can the team be formed if each developer can be assigned to only one project?

Solution: This question can be solved using the concept of combinations. We need to choose 6 developers out of 10 available developers, and the order in which they are chosen does not matter. The formula for calculating the number of combinations is:
nCr = n! / (r! * (n - r)!)
Where:
n = total number of items (in this case, developers)
r = number of items to be chosen (in this case, 6 developers)
! = factorial notation
Substituting the values in the formula, we get:
10C6 = 10! / (6! * (10 - 6)!)
= 10! / (6! * 4!)
= (10 * 9 * 8 * 7 * 6 * 5) / (6 * 5 * 4 * 3 * 2 * 1)
= 210
Therefore, the team of 6 developers can be formed in 210 different ways.

Question 2: A group of 8 friends is planning a trip and needs to arrange themselves in two cars, each with 4 seats. In how many ways can they arrange themselves in the cars?

Solution: This question can be solved using the concept of permutations and combinations. First, we need to choose 4 friends out of 8 to occupy the first car. This can be done in 8C4 ways, which is 8! / (4! * 4!) = 70 ways. Once we have chosen the 4 friends for the first car, we can arrange them in 4! = 24 ways in the car.
For the remaining 4 friends, there is only one way to arrange them in the second car (since all 4 seats are occupied).
Therefore, the total number of ways to arrange the 8 friends in the two cars is:
(8C4 * 4!) * 1 = (70 * 24) * 1 = 1,680 ways.

Question 3: A school has 15 different subjects, and a student needs to choose 5 of them for the upcoming semester. In how many ways can the student select the subjects?

Solution: This question can be solved using the concept of combinations. We need to choose 5 subjects out of 15 available subjects, and the order in which they are chosen does not matter. The formula for calculating the number of combinations is:
nCr = n! / (r! * (n - r)!)
Where:
n = total number of items (in this case, subjects)
r = number of items to be chosen (in this case, 5 subjects)
! = factorial notation
Substituting the values in the formula, we get:
15C5 = 15! / (5! * (15 - 5)!)
= 15! / (5! * 10!)
= (15 * 14 * 13 * 12 * 11) / (5 * 4 * 3 * 2 * 1)
= 3,003
Therefore, the student can select the 5 subjects in 3,003 different ways.

Question 4: A bookstore has 12 different textbooks, and a student wants to buy 4 of them for their courses. In how many different combinations can the student select the textbooks?

Solution: This question can be solved using the concept of combinations. We need to choose 4 textbooks out of 12 available textbooks, and the order in which they are chosen does not matter. The formula for calculating the number of combinations is:
nCr = n! / (r! * (n - r)!)
Where:
n = total number of items (in this case, textbooks)
r = number of items to be chosen (in this case, 4 textbooks)
! = factorial notation
Substituting the values in the formula, we get:
12C4 = 12! / (4! * (12 - 4)!)
= 12! / (4! * 8!)
= (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1)
= 495
Therefore, the student can select the 4 textbooks in 495 different combinations.

Question 5:  A team of 5 members needs to be selected from a group of 9 candidates for a project. Additionally, one member must be designated as the team leader. In how many ways can the team be selected and the leader chosen?

Solution: To choose the team leader, there are 9 candidates to choose from. Once the leader is chosen, there are 8 remaining candidates to choose the other 4 team members from. The number of ways to choose 4 from 8 is given by the combination formula: 8C4 = 8!/(4!(8-4)!) = 70. Therefore, the total number of ways to select the team and leader is 9 * 70 = 630.

Question 6: A school is organizing a science fair, and there are 6 different categories for students to enter their projects. If each student can only enter one category, in how many ways can 8 students choose their categories for the fair?

Solution: This is a problem of permutations with repetition. We are assigning 8 students to 6 categories, where repetition is allowed (multiple students can choose the same category). The number of ways to do this is 6^8 = 1,679,616.

Question 7: A group of 10 students is forming a committee to organize a charity event. The committee must consist of a chairperson, a secretary, and 3 other members. In how many ways can the committee positions be filled?

Solution: First, choose the chairperson from 10 students, which can be done in 10 ways. Then, choose the secretary from the remaining 9 students, which can be done in 9 ways. Finally, choose the 3 other members from the remaining 8 students, which can be done in 8C3 = 56 ways. The total number of ways is 10 * 9 * 56 = 5,040.

Question 8: A chef is planning a 4-course dinner menu, and there are 12 different dishes to choose from for each course. If the chef cannot repeat any dish in the same menu, in how many ways can the chef create the dinner menu?

Solution: For the first course, there are 12 dishes to choose from, so 12 ways. For the second course, there are 11 remaining dishes, so 11 ways. For the third course, there are 10 remaining dishes, so 10 ways. For the fourth course, there are 9 remaining dishes, so 9 ways. The total number of ways is 12 * 11 * 10 * 9 = 11,880.

Question 9: A company organizes a conference with 12 different workshops, and each participant can attend 5 workshops. In how many ways can a participant choose which workshops to attend?

Solution: To choose 5 workshops out of 12 different workshops, we need to calculate the combination of 12 things taken 5 at a time. The formula for combinations is nCr = n! / (r! * (n - r)!), where n is the total number of items, and r is the number of items to be chosen.
In this case, n = 12 (total number of workshops), and r = 5 (number of workshops to be chosen).
nCr = 12! / (5! * (12 - 5)!)
= 12! / (5! * 7!)
= 95,040
Therefore, a participant can choose which workshops to attend in 95,040 ways.

Question 10: A lottery game requires players to select 6 numbers from a pool of 49 numbers. In how many ways can a player choose their numbers for a ticket?

Solution: To select 6 numbers from a pool of 49 numbers, we need to calculate the combination of 49 things taken 6 at a time.
nCr = 49! / (6! * (49 - 6)!)
= 49! / (6! * 43!)
= 13,983,816
Therefore, a player can choose their numbers for a lottery ticket in 13,983,816 ways.

Question 11: A music playlist consists of 15 songs, and a listener wants to create a playlist of 10 songs without repeating any. In how many ways can the listener arrange the playlist?

Solution:  To arrange 10 songs out of 15 songs in a playlist without repeating any, we need to calculate the permutation of 15 things taken 10 at a time. The formula for permutations is nPr = n! / (n - r)!
In this case, n = 15 (total number of songs), and r = 10 (number of songs to be arranged).
nPr = 15! / (15 - 10)!
= 15! / 5!
= 1,307,504,000
Therefore, the listener can arrange the playlist in 1,307,504,000 ways.

Question 12: A chess tournament has 16 players. If each player plays against every other player exactly once, how many games will be played in total?

In a chess tournament with 16 players, each player plays against every other player exactly once. This means that the number of games played is equal to the combination of 16 things taken 2 at a time.
nCr = 16! / (2! * (16 - 2)!)
= 16! / (2! * 14!)
= 120
Therefore, a total of 120 games will be played in the tournament.

Question 13: A committee of 7 people needs to be formed from a group of 10 women and 8 men. If the committee must include at least 3 women, in how many ways can the committee be formed?

Solution: To form a committee of 7 people from a group of 10 women and 8 men, with the condition that the committee must include at least 3 women, we need to consider two cases:
Case 1: The committee has 3 women and 4 men.
Number of ways to choose 3 women from 10 women = 10C3 = 120
Number of ways to choose 4 men from 8 men = 8C4 = 70
Total number of ways = 120 * 70 = 8,400
Case 2: The committee has 4, 5, 6, or 7 women and the remaining members are men.
Number of ways to choose 4 women from 10 women = 10C4 = 210
Number of ways to choose 3 men from 8 men = 8C3 = 56
Total number of ways = 210 * 56 = 11,760
Number of ways to choose 5 women from 10 women = 10C5 = 252
Number of ways to choose 2 men from 8 men = 8C2 = 28
Total number of ways = 252 * 28 = 7,056
Number of ways to choose 6 women from 10 women = 10C6 = 210
Number of ways to choose 1 man from 8 men = 8C1 = 8
Total number of ways = 210 * 8 = 1,680
Number of ways to choose 7 women from 10 women = 10C7 = 120
Number of ways to choose 0 men from 8 men = 1 (no choice)
Total number of ways = 120 * 1 = 120
Total number of ways for the committee formation = 8,400 + 11,760 + 7,056 + 1,680 + 120 = 29,016
Therefore, the committee can be formed in 29,016 ways.

Question 14: A classroom has 25 students, and the teacher wants to form groups of 5 for a project. In how many ways can the teacher divide the students into groups?

Solution: To divide 25 students into groups of 5, we need to calculate the combination of 25 things taken 5 at a time, and then divide the result by the permutation of 5 things taken 5 at a time (since the order of students within a group doesn't matter).
nCr = 25! / (5! * (25 - 5)!)
= 25! / (5! * 20!)
= 53,130
Number of ways to arrange 5 students within a group = 5!
Total number of ways to divide the students into groups = 53,130 / 5!
= 142,506
Therefore, the teacher can divide the students into groups in 142,506 ways.

Question 15: A box contains 8 different types of chocolates, and a person wants to create a gift box with 4 chocolates. In how many ways can the person select the chocolates for the gift box?

Solution: To select 4 chocolates from 8 different types of chocolates, we need to calculate the combination of 8 things taken 4 at a time.
nCr = 8! / (4! * (8 - 4)!)
= 8! / (4! * 4!)
= 70
Therefore, the person can select the chocolates for the gift box in 70 ways.

Question 16: A bookstore has 20 different novels by various authors. If a customer wants to buy 3 novels written by the same author, in how many ways can the customer select the novels?

Solution: To select 3 novels written by the same author from a collection of 20 different novels, we need to consider two steps:
Step 1: Choose the author.
Number of ways to choose the author = 20 (assuming all novels are written by different authors)
Step 2: Choose 3 novels from the selected author.
Number of ways to choose 3 novels from the same author = 3C3 = 1
Total number of ways = Number of ways to choose the author * Number of ways to choose 3 novels from the selected author
= 20 * 1
= 20
Therefore, the customer can select the novels in 20 ways.

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