# a committee of 3 women and 4 men is to be formed from 6 women and 7 men [duplicate]

a committee of 3 women and 4 men is to be formed from 6 women and 7 men. How many ways committee can be formed if it can only include at most 1 of the youngest woman or youngest man.

case 1: neither are included -> 5C3 X 6C2

case 2: only youngest woman included -> 5C3 X 7C4

case 3: only youngest man included -> 6C3 X 6C3

i then add all of those up together. is this right?

## marked as duplicate by lulu, N. F. Taussig combinatorics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 27 '18 at 12:17

• All three are wrong. For $1$ you are choosing five people, for $2$ some of your combinations include the youngest man, for $3$ some of your combinations include the youngest woman. – lulu Nov 27 '18 at 11:37
Without the age restriction, there would be $$\binom{6}{3}\binom{7}{4}$$. Of these, exactly $$\binom{5}{2}\binom{6}{3}$$ are illegal due to featuring the youngest member of each sex. This makes the number of legal solutions $$\binom{6}{3}(\binom{7}{4}-\binom{5}{2})=20\times (35-10)=500$$.