1. Applications of Permutations and Combinations
Certain combinatorics problems are harder to solve and require some ingenuity. One gets to be good at solving such problems by applying clear logic and practice.

2. Example 1:
Find the number of ways of arranging 6 women and 3 men to stand in a row, so that all 3 men are standing together.
This problem is simplified by treating the men as a single unit. One approach might be:
𝑊 1 𝑊 2 𝑊 5 𝑊 6 𝑊 4 𝑊 𝑀 1 𝑀 3 𝑀 2
where 𝑊 𝑖 represents women and 𝑀 𝑖 represents men.
The number of permutations of these 7 ‘objects’ is 7! However, the men can be arranged in 3! ways, and so the total number of ways to arrange the 6 women and 3 men so that all 3 men are standing together is: 7!×3!=30240

3. Example 2:
Find the number of ways of arranging 6 women and 3 men in a row so that no two men are standing next to one another.
This problem can be solved by arranging the women with a space in between them and also a space before the first woman, and a space after the last woman:
___ 𝑊 1 ___ 𝑊 2 ___ 𝑊 5 ___ 𝑊 6 ___𝑊 4 ___𝑊 3 ___
There are 6! possible arrangements of the 6 women. The remaining 3 men can be arranged in 7×6×5 different ways, giving a total of
6!×7×6×5= different ways of arranging them according to the required conditions.

4. P each couple stands together = 6!× (2!) 6 12! = 1 10395
Example 3:
A group of 12 people consisting of 6 married couples is arranged at random in a line for a photograph. Find the probability that each wife is standing next to her husband.
If there are no restrictions, the number of arrangements is 12! However, we can treat the husband and wife as a unit, in which case the total number of arrangements is 6! But each couple can be arranged in 2! ways. So that gives us 6!× (2!) 6 (*) ways of arranging the couples. Thus the probability is given by:
P each couple stands together = 6!× (2!) 6 12! =
= =9.62× 10 −5
(*) To understand (2!) 6 , think of tree diagram.

5. Example 4:
Four letters are to be selected from the letters in the word RIGIDITY. How many different combinations are there?
We have to consider the cases where there are combinations with no 𝐼s, one 𝐼, two 𝐼s and three 𝐼s.
Zero 𝐼s : =5
One 𝐼 : =10
Two 𝐼s : =10
Three 𝐼s : =5
The total number of combinations is therefore the sum of all the above: =30.
What do you notice about these combinations?

7. The problem can be solved easily by noticing that the points in the spider’s web correspond to binomial coefficients.
Every path consists of exactly 7 moves. The problem is solved once we can determine how many such moves are possible.
To arrive at (4,3), the spider must move 4 across and 3 up in any one of the possible ways to reach the fly.
Next slide shows solution.

8. Therefore, the total number of ways the spider can reach the fly is 35.

9. Also do miscellaneous exercise 5.
Now do Exercise 5C .
Also do miscellaneous exercise 5. 