Permutation of Groups into Rows I

10 chairs are numbered and arranged in 2 rows as shown in the following diagram. 

In how many ways can 9 people, comprising 7 men and 2 women, be seated

(i) without restriction?

(ii) if the 2 women must not sit in the same row?

(iii) if the 2 women must sit together in the same row?

Later, these 9 people are split into 3 groups of 3 people each. In how many ways can this be done? 

Solutions

(i) 10! = 3628800

Note: that an empty seat is a distinct object, treat the empty seat as another person. 

(ii) 2016000

women: 2 vs non-women: 8

 

Total arrangement: 2C1 x 8C4 x 5! x 5! = 2016000

note: 

choosing 1 woman to first row: 2C1

choosing non-women to first row: 8C4

No. of arrangement in 1st row: 5!

No of 2nd row arrangements: 5!

(iii) 645120

 Case I (2 women first row) and Case II (2 women second row)

 

Total arrangements

= 2C2 x 8C3 x 4! x 2! x 5! x 2!

= 645120

Note: 

Choosing non-women to the first row: 8C3

Arrangements for first row: 4! x 2! 

Arrangements for second row: 5!

2 cases: 2

 

Splitting them into 3 groups of 3 people

Choosing 9 people into the first group: 9C3

Choosing remaining 6 people into the second group: 6C3

Choosing the remaining 3 people into the 3rd group: 3C3

Three groups of the same size: divide by 3!

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