Question

Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements Select one: a. None of these b. 1440 c. 720 d. 360

Answer 1

Answer:

1440

Step-by-step explanation:

First women can take any of the chairs marked 1 to 4 in 4 different way.

Second women can take any of the remaining 3 chairs from those marked 1 to 4 in 3 different ways.

So,total no of ways in which women can take seat =4×3

⇒4P2

4P2=4!(4−2)!

=4×3×2×12×1

=12

After two women are seated 6 chairs remains

First man take seat in any of the 6 chairs in 6 different ways,second man can take seat in any of the remaining 5 chairs in 5 different ways

Third man can take seat in any of the remaining 4 chairs in 4 different ways.

So,total no of ways in which men can take seat =6×5×4

⇒6P3

6P3=6!(6−3)!

⇒6×5×4×3×2×13×2×1

⇒120

Hence total number of ways in which men and women can be seated =120×12

⇒1440

Answer 2

Answer:

The correct answer will be -

(b) 1440

Step-by-step explanation:

Given,

First, the women choose the chairs amongst the chairs numbered 1 to 4. It will be in 4 ways.

Now the second woman will choose out of 3 chairs in 3 ways.

So total ways in which women can take seats = $4\times 3 =$ 12 ways

Now, there are 6 chairs left and 3 men wish to occupy one chair each.

This can be done in - $^6P_3$ ways like the total permutation of selecting 3 items out 6 items

$^6P_3 = \frac{6!}{(6-3)!} = 6\times5\times4 = 120 ways$

So total ways so that everyone seats will be = $12\times 120$ = 1440 ways.

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