# CRE 6 - Seating arrangements | Modern Math - Permutation & Combination

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**CRE 6 - Seating arrangements | Modern Math - Permutation & Combination**

**Answer the next 10 questions based on the information given below:**

In how many ways can 16 people be seated

- (a)
16! - 1

- (b)
15!

- (c)
16!

- (d)
None of these

Answer: Option C

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**Explanation** :

16 people can be seated in a row in 16! ways.

Hence, option (c).

Workspace:

**CRE 6 - Seating arrangements | Modern Math - Permutation & Combination**

On a circular table having 16 chairs?

- (a)
16! - 1

- (b)
15!

- (c)
16!

- (d)
None of these

Answer: Option B

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**Explanation** :

Number of ways of seating n people around a circular table is (n – 1)!.

Since initially all the chairs are similar, the first person can be seated in only 1 way.

Once the first person is seated, all other 15 chairs are different, hence 15 people can be seated in 15! ways.

Hence the required answer = (16 – 1)! = 15!

Hence, option (b).

Workspace:

**CRE 6 - Seating arrangements | Modern Math - Permutation & Combination**

On a circular table having 15 blue chairs & 1 green chair?

- (a)
16! - 1

- (b)
15!

- (c)
16!

- (d)
None of these

Answer: Option C

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**Explanation** :

Since here all the chairs around the table are not the same initially because of 1 green chair.

16 people are to be seated on 16 different chairs, no. of ways = 16!

Hence, option (c).

Workspace:

**CRE 6 - Seating arrangements | Modern Math - Permutation & Combination**

On a rectangular table having 5 chairs each on the longer sides and 3 chairs each on the shorter sides?

- (a)
8 × 15!

- (b)
15!

- (c)
16!

- (d)
4 × 15!

Answer: Option A

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**Explanation** :

Here, initially seats 1 and 9 are similar.

Similarly, from the figure, there are 8 different pair of seats i.e., 1 & 9, 2 & 10, 3 & 11, 4 & 12, 5 & 13, 6 & 14, 7 & 15 and 8 & 16.

Hence, the first person can be seated in 8 different ways on one seat out of the 8 different pairs of seats.

Once the first person is seated, the remaining 15 seats become different from each other.

∴ 15 people can be seated on these 15 different chairs in 15!

⇒ Total number of ways = 8 × 15!

Hence option (a).

Workspace:

**CRE 6 - Seating arrangements | Modern Math - Permutation & Combination**

On a square table having 4 chairs on each side?

- (a)
15!

- (b)
16!

- (c)
8 × 15!

- (d)
4 × 15!

Answer: Option D

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**Explanation** :

Here, initially seats 1, 5, 9 and 13 are similar.

Similarly, from the figure, there are 4 different pair of seats i.e., (1, 5, 9, 13), (2, 6, 10, 14), (3, 7, 11, 15) and (4, 8, 12, 16).

Hence, the first person can be seated in 4 different ways on one seat out of the 4 different pairs of seats.

Once the first person is seated, the remaining 15 seats become different from each other.

∴ 15 people can be seated on these 15 different chairs in 15!

⇒ Total number of ways = 4 × 15!

Hence option (d).

Workspace:

**CRE 6 - Seating arrangements | Modern Math - Permutation & Combination**

1 circular table with 9 chairs & rest 7 in a row?

- (a)
8! × 7!

- (b)
^{16}C_{7}× 7! × 8! - (c)
^{16}C_{7}× 9! × 7! - (d)
None of these

Answer: Option B

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**Explanation** :

To seat 7 people in a row, we first need to select 7 people out of 16.

Number of ways of selecting 7 people out of 16 = ^{16}C_{7}.

Number of ways of arranging these 7 people in a row = 7!.

∴ Total number of ways of seating 7 people in a row out of 16 = ^{16}C_{7} × 7!

Remaining 9 people can be seated around a circular table in (9 – 1)! = 8!.

∴ Total number of ways of seating 16 people around a circular table with 9 chairs & rest 7 in a row = ^{16}C_{7} × 7! × 8!

Hence, option (b).

Workspace:

**CRE 6 - Seating arrangements | Modern Math - Permutation & Combination**

2 circular tables one with 10 chairs & other with 6 chairs?

- (a)
^{16}C_{10}× 9! × 5! - (b)
^{16}C_{10}× 10! × 6! - (c)
10! × 6!

- (d)
None of these

Answer: Option A

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**Explanation** :

**Table 1 with 10 chairs:**

Number of ways of selecting 10 people out of 16 = ^{16}C_{10}.

Number of ways of arranging these 10 people around a circle = 9!

∴ Total number of ways of seating 10 people around a circle = ^{16}C_{10} × 9!

**Table 2 with 6 chairs:**

Remaining 6 people can be seated around the 2nd table in 5! ways.

∴ ∴ Total number of ways of seating 16 people around 2 circular tables one with 10 chairs & other with 6 chairs = ^{16}C_{10} × 9! × 5!

Hence, option (a).

Workspace:

**CRE 6 - Seating arrangements | Modern Math - Permutation & Combination**

4 different circular tables each with 4 chairs?

- (a)
$16!\times {\left(\frac{3!}{4!}\right)}^{4}$

- (b)
(3!)

^{4} - (c)
(4!)

^{4} - (d)
None of these

Answer: Option A

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**Explanation** :

**Table 1:** First we select 4 people and then seat them on the table.

Number of ways of selecting 4 people out of 16 = ^{16}C_{4}

Number of ways of seating 4 people around the table = (4 – 1)! = 3!

∴ Total number of ways of seating 4 people around table 1 = ^{16}C_{4} × 3!

**Table 2:** First we select 4 people and then seat them on the table.

Number of ways of selecting 4 people out of remaining 12 = ^{12}C_{4}

Number of ways of seating 4 people around the table = (4 – 1)! = 3!

∴ Total number of ways of seating 4 people around table 2 = ^{12}C_{4} × 3!

**Table 3:** First we select 4 people and then seat them on the table.

Number of ways of selecting 4 people out of remaining 8 = ^{8}C_{4}

Number of ways of seating 4 people around the table = (4 – 1)! = 3!

∴ Total number of ways of seating 4 people around table 4 = ^{8}C_{4} × 3!

**Table 4:** Now we have only 4 people remaining.

∴ Total number of ways of seating 4 people around table 4 = 3!

∴ Total number of ways of seating 16 people around the 4 different tables = ^{16}C_{4} × ^{12}C_{4} × ^{8}C_{4} × 3! × 3! × 3! × 3!

= $\frac{16!}{4!\times 12!}\times \frac{12!}{4!\times 8!}\times \frac{8!}{4!\times 4!}$ × 3! × 3! × 3! × 3!

= 16! × ${\left(\frac{3!}{4!}\right)}^{4}$

Hence, option (a).

Workspace:

**CRE 6 - Seating arrangements | Modern Math - Permutation & Combination**

4 identical circular tables each with 4 chairs?

- (a)
$16!\times {\left(\frac{3!}{4!}\right)}^{4}$

- (b)
$16!\times \frac{{3!}^{4}}{{4!}^{5}}$

- (c)
(3!)

^{3} - (d)
None of these

Answer: Option B

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**Explanation** :

Consider the solution to previous question.

Total number of ways of seating 16 people around the 4 tables = 16! × ${\left(\frac{3!}{4!}\right)}^{4}$

Now, when all 4 tables are similar, it would not matter a group of 4 people sit on which table.

So, 4 group of people can be distributed on 4 different tables in 4! ways. But, since all tables are similar 4 group of people can be distributed on 4 similar tables in only 1 way.

∴ Number of ways from previous question will get divided by 4!.

⇒ Required answer = $16!\times {\left(\frac{3!}{4!}\right)}^{4}\xf74!=16!\times \frac{{3!}^{4}}{{4!}^{5}}$

Hence, option (b).

Workspace:

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