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

Answer: Option C

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Explanation :

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

Hence, option (c).

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).

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).

Answer: Option A

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Explanation :

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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).

Answer: Option D

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Explanation :

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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).

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 = ¹⁶C₇.

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 = ¹⁶C₇ × 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 = ¹⁶C₇ × 7! × 8!

Hence, option (b).

Answer: Option A

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Explanation :

Table 1 with 10 chairs:

Number of ways of selecting 10 people out of 16 = ¹⁶C₁₀.

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

∴ Total number of ways of seating 10 people around a circle = ¹⁶C₁₀ × 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 = ¹⁶C₁₀ × 9! × 5!

Hence, option (a).

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 = ¹⁶C₄

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

∴ Total number of ways of seating 4 people around table 1 = ¹⁶C₄ × 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 = ¹²C₄

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

∴ Total number of ways of seating 4 people around table 2 = ¹²C₄ × 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 = ⁸C₄

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

∴ Total number of ways of seating 4 people around table 4 = ⁸C₄ × 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 = ¹⁶C₄ × ¹²C₄ × ⁸C₄ × 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).

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 = $16!\times {\left(\frac{3!}{4!}\right)}^4÷4!=16!\times \frac{{3!}^4}{{4!}^5}$     

Hence, option (b).