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 ?Options3660132401440

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

Step 1: Select 2 chairs from the first 4 chairs for the women. This can be done in 4C2 ways. The formula for combinations is nCr = n! / r!(n-r)!. Here, n is the total number of items, and r is the number of items to choose. So, 4C2 = 4! / 2!(4-2)! = 6 ways.

Step 2: Select 3 chairs from the remaining 6 chairs for the men. This can be done in 6C3 ways. So, 6C3 = 6! / 3!(6-3)! = 20 ways.

Step 3: Multiply the number of ways from step 1 and step 2 to get the total number of ways. So, 6 * 20 = 120 ways.

Therefore, the maximum number of different ways in which this can be done is 120.

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

Step 1: We first select 2 chairs from the 4 available 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 from the remaining 6 for the men. Using the same formula, n=6 (chairs) and r=3 (men), so C(6, 3) = 6! / [(6-3)!3!] = 20.

Step 3: Since the women's and men's chair selections are independent events, we multiply the number of ways they can occur to get the total number of ways both can occur. So, the total number of ways is 6 * 20 = 120.

Therefore, the maximum number of different ways in which this can be done is 120.