How many ways can a committee be formed from 4 men and 6 women with... 4 members, at least two whom are women and Mr and Mrs Baggins cannot be chosen
Any help is appreciated!!!!! #:D Thanks Joe!!!!!
On
$2$ women committee - $\binom{6}{2}\cdot \binom{4}{2} - (3\cdot 5) = 75$
Explanation: Pick $2$ from $6$ women $(15)$ times $2$ from $4$ men $(6)$ minus the number of combinations which include Mr and Mrs Baggins $(15)$. Mrs Baggins can be paired with $5$ other women, and for every one of those, Mr Baggins can be paired with $3$ other men.
$3$ women committee - $\binom{6}{3}\cdot \binom{4}{1} - 10 = 70$
In this case, there are $\binom{5}{2} = 10$ ways that Mrs Baggins can be grouped with $2$ other women and Mr Baggins.
$4$ women committee - $\binom{6}{4} = 15$
Total $= 160$
On
Here is my idea behind this...
We need at least two women and mr and mrs Baggins cannot be included.
Case 1: 2 women
(4C2)(6C2)
But Mr and Mrs Baggins can be in this arrangement so we need to subtract them out they appear in this arrangement 15 times so subtract that out
Case 2: 3 Women (4C1)(6C3) Both Mr and Ms Baggins can be in this arrangement so we need to subtract them out they appear in this arrangement 10 times so subtract that out
Case 3: 4 Women (4C0)(6C4) Only Ms Baggins is included in this arrangement
So we are left with 75+70+15 which gives you 160 ways
You can have 2 men and 2 women, or 1 man and 3 women, or just 4 women. If we temporarily ignore the Baggins business, that $${4\choose2}{6\choose2}+{4\choose1}{6\choose3}+{6\choose4}$$
Now of the 2 men, 2 women ways, $3\times5$ use both Bagginses; of the 1 man, 3 women ways, ${5\choose2}$ ways use both Bagginses, and we're done.