How to do 191st question?

• When there is one lady, there is five gentleman in the committee. The ways to form the committee is `color(red)((""^6C_5xx""^5C_1)=30`.
• When there is two lady, there is four gentleman in the committee. The ways to form the committee is `color(red)((""^6C_4xx""^5C_2)= 150`.
• When there is three lady, there is three gentleman in the committee. The ways to form the committee is `color(red)((""^6C_3xx""^5C_3)=200`.
• When there is four lady, there is two gentleman in the committee. The ways to form the committee is `color(red)((""^6C_2xx""^5C_4)=75`.
• When there is five lady, there is one gentleman in the committee. The ways to form the committee is `color(red)((""^6C_1xx""^5C_5)=6`.

The sum of the ways is `color(blue)(30+150+200+75+6)=461`

That is my answer, I don't know why the option doesn't match.

`6` gentlemen, `5` ladies form `6`-person committee, which at least

Total number of possible outcomes,

`"^11C_6`

Total number of unconditioned outcomes (`6` gentlemen, `0` lady),

`""^6C_6 *"^5C_0`

Total number of conditioned outcomes,

`""^11C_6-(""^6C_6 *"^5C_0)`

Key in your calculator,

`461`