We need to select 4 men (M) and 4 women (W) from the two groups. Consider the following cases:
\(\text{From Group A}\) | \(\text{From Group B}\) | \(\text{Ways of Selection}\) |
---|---|---|
4M | 4W | \({{4}\choose{4}} \cdot {{4}\choose{4}} = 1\) |
3M1W | 1M3W | \({{4}\choose{3}} \cdot {{5}\choose{1}} \cdot {{5}\choose{3}} \cdot {{4}\choose{1}} = 400\) |
2M2W | 2M2W | \({{4}\choose{2}} \cdot {{5}\choose{2}} \cdot {{5}\choose{2}} \cdot {{4}\choose{2}} = 3600\) |
1M3W | 3M1W | \({{4}\choose{1}} \cdot {{5}\choose{3}} \cdot {{5}\choose{1}} \cdot {{4}\choose{3}} = 1600\) |
4W | 4M | \({{5}\choose{4}} \cdot {{5}\choose{4}} = 25\) |
Total | 5626 |
Final Answer: 5626.
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)