In how many ways 2 directors and 3 executives can be arranged for a meeting? If there are 6 chairs available two on one side and remaining four on the other side of the table and the two directors has to be together on one side and the executives on the other side.a)24b)48c)720d)120Correct answer is option 'B'. Can you explain this answer?

Question

Answer

3 executives in 4 chairs = 4P3 ways
Total no. of ways = 2! * 4P3
=> 2*1*(4!/1!)
= 2*24
= 48

Answer

Total number of ways to arrange 2 directors and 3 executives = 5! = 120.

But, the two directors have to be together on one side.

So, we can consider them as a single unit and arrange them in 2! ways among themselves.

Also, the 3 executives have to be arranged on the other side.

So, we can arrange them in 3! ways among themselves.

Therefore, the total number of ways to arrange 2 directors and 3 executives with the given condition = 2! × 3! = 12.

Now, we have to arrange these 12 arrangements on the two sides of the table.

The two directors can sit on either of the two chairs on one side of the table in 2! ways.

Therefore, the total number of ways to arrange 2 directors and 3 executives with the given condition and given seating arrangement = 12 × 2! = 24.

Hence, the correct option is (B) 48.

Answer

The directors can be seated in 2 ways(2!=2)
For each way in which the directors can seat ----> no. of places for executives(n) = 4 no. of executives (r) = 3 so, no.of ways of sitting executives = 4!/(4-3)! = 24 so, total no. of ways = 24× 2 = 48

Answer

Answer

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