Each of two women and three men is to occupy one chair out of eight chairs, each of which is numbered from one to eight. First , women are to occupy any two chairs from those numbered one to four, and then the three men would occupy any three chairs out of the remaining six chairs. What is the maximum number of different ways in which this can be done ?Options4014403660132

The problem can be solved by using the concept of combinations in probability.

Step 1: We first select 2 chairs out of 4 for the women. The number of ways to do this is given by the combination formula C(n, r) = n! / [(n-r)!r!], where n is the total number of items, and r is the number of items to choose. Here, n=4 (chairs) and r=2 (women). So, C(4, 2) = 4! / [(4-2)!2!] = 6.

Step 2: We then select 3 chairs out of the remaining 6 for the men. Using the same formula, we find C(6, 3) = 6! / [(6-3)!3!] = 20.

Step 3: Since the women and men can sit in any arrangement on their selected chairs, we multiply the number of chair selections by the number of arrangements. The number of arrangements is given by the permutation formula P(n) = n!, where n is the number of items. So, P(2) = 2! = 2 for the women and P(3) = 3! = 6 for the men.

Step 4: The total number of ways is therefore (number of chair selections for women) * (arrangements for women) * (number of chair selections for men) * (arrangements for men) = 6 * 2 * 20 * 6 = 1440.

So, the maximum number of different ways in which this can be done is 1440.