PERMUTATION COMBINATION 11th Class MH Board

                          EXERCISE 3.5 


1.In how many different ways can 8 friends seat around a table?

Answer:

We know that ‘n’ persons can sit around a table in (n − 1)! ways

∴ 8 friends can sit around a table in 7! ways
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040 ways

∴ 8 friends can sit around a table in 5040 ways.

2. A party has 20 participants. Find the number of distinct ways for the host to sit with them around a circular table. How many of these ways have to specified persons on either side of the host?

Answer

A party has 20 participants.

All of them and the host (i.e., 21 persons) can be seated at a circular table in (21 - 1)! = 20! ways.

When two particular participants are seated on either side of the host.
Host takes the chair in 1 way.
These 2 persons can sit on either side of the host in 2! ways

Once the host occupies his chair, it is not a circular permutation any more.
Remaining 18 people occupy their chairs in 18! ways.

∴ Total number of arrangements possible if two particular participants are seated on either side of the host = 2! × 18!

3. Delegates From 24 countries participate in a round table discussion. Find the number of sitting arrangements were to specified delegates are 
a)they always together 
b) never together.

Answer-

Delegates of 24 countries are to participate in a round table discussion such that two specified delegates are always together.

Let us consider these 2 delegates as one unit.
They can be arranged among themselves in 2! ways.

Also, these two delegates are to be seated with 22 other delegates (i.e. total 23) which can be done in (23 − 1)! = 22! ways

∴ Total number of arrangement if two specified delegates are always together
= 22! × 2!
b)When 2 specified delegates are never together then, the other 22 delegates can participate in a round table discussion in (22  1)! = 21! ways.

∴ There are 22 places of which any 2 places can be filled by those 2 delegates who are never together.

∴ Two specified delegates can be arranged in 

∴ Total number of arrangements if two specified delegates are never together
`=^{22}P_2×21!`
`=\frac{\text{22!}}{\text{(22-2)!}×21!`
`=\frac{\text{22!}}{\text{20!}}×21!`
`=22×21×21!`
`=21×22!`

4. Find the number of ways for 15 people to sit around the table so that no two arrangements have the same neighbors . 

Answer-

15 people can sit around a table in (15 – 1)! = 14! ways.
Total number of arrangements = 14!

Now, the number of arrangements in which any person can have the same neighbors on either side by clockwise or anticlockwise arrangements = `\frac{\text{14!}}{\text{2!}}`

∴ The number of arrangements in which no two arrangements have the same neighbors
`=14!-\frac{\text{14!}}{\text{2!}}`
`=14!-\left(1-\frac{\text{1}}{\text{2}}\right)`
 `=14!×\frac{\text{1}}{\text{2}}`
`=\frac{\text{14!}}{\text{2!}}`

5. A committee of 10 members seeds around the table find the number of arrangements that have the president and the vice president together.

Answer-

A committee of 20 members sits around a table.
But, President and Vice-president sit together.

Let us consider the President and Vice-president as one unit.
They can be arranged among themselves in 2! ways.

Now, this unit with the other 18 members of the committee is to be arranged around a table, which can be done in (19 − 1)! = 18! ways.

∴ Total number of arrangements possible if President and Vice-president sit together
 = 18! × 2!

6. 5 men 2 women and a child sit around a table. Find the number of arrangements where the child is seated 
a)between the two women
b) between two men.

Answer-

a)Five men, two women, and a child sit around a table.
Child is seated between the two women.

∴ The two women can be seated on either side of the child in 2! ways.
Let us consider these 3 (two women and a child) as one unit.

Also, these 3 are to seated with 5 men,
(i.e. total 6 units) which can be done in (6 − 1)! = 5! ways

∴Total number of arrangements if the child is seated between two women = 5! × 2!

b)Five men, two women, and a child sit around a table.
Here, the child is seated between 2 men.
Since there are 5 men, such a group can be formed in `^5P_2`
`=\frac{\text{5!}}{\text{3!}}`
= 5 × 4
= 20 ways

Thus, there are 20 ways in which the child can be seated between 2 men.
We consider the 2 men and the child between as one unit.
Also, we have 3 more men and 2 women.
Thus, we have 1 + 3 + 2 = 6 persons.

These 6 persons can be arranged around a table in (6 – 1)! = 5! – ways.

∴ Total number of required arrangements
= 20 × 5!
= 20 × 120
= 2400

 7. Eight men and Six women sit around a table. How many of sitting arrangements will have known to woman 2 whether? 

Answer-

8 men can be seated around a table in (8 − 1)! = 7! ways.
There are 8 gaps created by 8 men’s seats.

∴ 6 Women can be seated in 8 gaps in `^8P_6` ways

∴ Total number of arrangements so that no two women are together 
`= 7! × ^8P_6`

8. Find the number of sitting arrangements for the three men and 3 women to sit around a table so that exactly two women are together?

Answer-

Two women sit together and one woman sits separately.

Woman sitting separately can be selected in 3 ways.

Other two women occupy two chairs in one way (as it is circular arrangement). They can be seated on those two chairs in 2 ways. Suppose two chairs are chairs 1 and 2 shown in the figure. Then the third woman has only two options viz chairs 4 or 5.

∴ Third woman can be seated in 2 ways. 3 men are seated in 3! ways
Required number = 3 × 2 × 2 × 3!

9. Four objects in a set of 10 objects are alike. Find the number of ways of arranging them in a circular order. 

Answer -

There are 10 objects.
These 10 objects can be arranged in a circular order in (10 – 1)! = 9! ways.
∴ n = 9!

Out of 10 objects, 4 are alike.
r = 4

∴ Required number of arrangements
`=\frac{\text{n!}}{\text{r!}}`
`=\frac{\text{9!}}{\text{4!}}`

10. 15 persons sit around a table find the number of arrangements that have to specified person not sitting side by side.
Answer-
Since 2 particular persons can’t be sitting side by side.

The other 13 persons can be arranged around the table in (13 − 1)! = 12!

13 people around a table create 13 gaps in which 2 person are to be seated
Number of arrangements of 2 person = `^{13}P_2`

∴ Total number of arrangements in which two specified persons not sitting side by side
= 12! × $^{13}P_2$
= 12 × 13 × 12
= 13 × 12! × 12
= 12 × 13!


PERMUTATION COMBINATION EXERCISE 3.5 11th Class MH Board
PERMUTATION COMBINATION EXERCISE 3.5 11th Class MH Board

Permutation Combination Exercise 3.5