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#