It would be 46, have a nice day
Given:
Consider the equation is:

To prove:
by using the properties of logarithms.
Solution:
We have,

Taking left hand side (LHS), we get

![\left[\because \log_ab=\dfrac{\log_x a}{\log_x b}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbecause%20%5Clog_ab%3D%5Cdfrac%7B%5Clog_x%20a%7D%7B%5Clog_x%20b%7D%5Cright%5D)

![[\because \log x^n=n\log x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Clog%20x%5En%3Dn%5Clog%20x%5D)

![\left[\because \log_ab=\dfrac{\log_x a}{\log_x b}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbecause%20%5Clog_ab%3D%5Cdfrac%7B%5Clog_x%20a%7D%7B%5Clog_x%20b%7D%5Cright%5D)

Hence proved.
(a) there are 8C2 = 28 ways of picking 2 girls from 8
And there are 21C4 = 5985 ways of picking 4 boys
Required number of ways for 2g / 4b = 28 * 5985 = 167,580
(b) at least 2 girls means combinations of 2g/4b , 3g,3b , 4g/2b , 5g 1b or
6 girls.
2g/4b = 167,580 ways
3g/3b = 8C3 * 21C3 = 56 * 1330 = 74,480
4g/2b = 8C4* 21C2 = 70 * 210 = 14,700
5g 1b = 8C5* 21 = 56*21 = 1176
6 girls = 8C6 = 28
adding these up we get the answer to (b) which is 257,964
It's easy it's just 62×9 it equals 558