MAT prob&stat1 permutations and combinations

Question 1

What is the factorial function and its significance in arrangements?

Answer 1

Factorial (n!) represents the number of ways to arrange n distinct objects.
n! = n × (n - 1) × (n - 2) × … × 1.

Question 2

Example: How many different 5-digit numbers can be made from {1, 2, 3, 4, 5} if:
a) Digits are not repeated.
b) Digits can be reused.

Answer 2

a) 5! = 120
b) 5^5 = 3125

Question 3

How do you arrange objects with repeated elements?

Answer 3

For n objects where some elements are repeated, divide by the factorial of the repetitions.
Number of arrangements = n! / (p1! × p2! × …).

Question 4

Example: How many ways can the letters of “PUPPY” be arranged?

Answer 4

5! / 3! = 120 / 6 = 20 (since “P” repeats 3 times).

Question 5

What is a permutation?

Answer 5

A permutation is an arrangement of objects where order matters.
Number of permutations: P(n, r) = n! / (n - r)!

Question 6

Example: A club of 25 members needs 5 officials. How many ways can they be chosen?

Answer 6

P(25, 5) = 25! / 20! = 25 × 24 × 23 × 22 × 21 = 6,375,600

Question 7

What is a combination?

Answer 7

A combination selects objects where order does not matter.
Number of combinations: C(n, r) = n! / [(n - r)! × r!].

Question 8

Example: A committee of 5 is chosen from 25 members. How many ways can this be done?

Answer 8

C(25, 5) = 25! / [20! × 5!] = 53,130

Question 9

How do you handle selection problems with restrictions?

Answer 9

1. Count favorable arrangements directly or by subtraction.
2. Apply combinations or permutations as needed.

Question 10

Example: How many ways can 6 people (A, B, C, D, E, F) line up if A and B cannot stand next to each other?

Answer 10

Total arrangements: 6! = 720.
Arrangements with A and B together: 5! × 2 = 240.
Required arrangements: 720 - 240 = 480.

Question 11

Example: I have a box of chocolates with 10 different chocolates left in it. Of these, there are 6 which I particularly like. However, I intend to offer my three friends one chocolate each before I eat the rest. What is the probability that I will be left with at least 5 chocolates that I particularly like?

Answer 11

Total ways to choose 3 chocolates: C(10, 3) = 120.
Ways where none of the liked chocolates are chosen: C(4, 3) = 4.
Ways where one liked chocolate is chosen: C(6, 1) × C(4, 2) = 6 × 6 = 36.
Favorable outcomes: 4 + 36 = 40.
Probability = 40 / 120 = 1/3.