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Question Number 5280 by Rasheed Soomro last updated on 04/May/16

$$\mathrm{A}\mathrm{delegation}\mathrm{of}4\mathrm{people}\mathrm{is}\mathrm{to}\mathrm{be}$$$$\mathrm{selected}\mathrm{from}5\mathrm{women}\mathrm{and}6\mathrm{men}.$$$$\mathrm{Find}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{delegations}\mathrm{if}$$$$\left(\mathbf{a}\right)\mathrm{there}\mathrm{are}\mathrm{no}\mathrm{restrictions},$$$$\left(\mathbf{b}\right)\mathrm{there}\mathrm{is}\mathrm{at}\mathrm{least}1\mathrm{woman},$$$$\left(\mathbf{c}\right)\mathrm{there}\mathrm{are}\mathrm{at}\mathrm{least}2\mathrm{women}.$$$$\mathrm{One}\mathrm{of}\mathrm{the}\mathrm{men}\mathrm{cannot}\mathrm{get}\mathrm{along}\mathrm{with}$$$$\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{women}.\mathrm{Find}\mathrm{the}\mathrm{number}\mathrm{of}$$$$\mathrm{delegations}\mathrm{which}\mathrm{include}\mathrm{this}\mathrm{particular}$$$$\mathrm{man}\mathrm{or}\mathrm{woman},\mathrm{but}\mathrm{not}\mathrm{both}.$$$$$$

Answered by Yozzii last updated on 04/May/16

$$\left(\mathbf{a}\right)\mathrm{no}.\mathrm{of}\mathrm{ways}=\left(\begin{array}{c}5+6\\ 4\end{array}\right)=\left(\begin{array}{c}11\\ 4\end{array}\right)=330$$$$$$$$\left(\mathbf{b}\right)\mathrm{no}.\mathrm{of}\mathrm{ways}=\mathrm{total}\mathrm{no}.\mathrm{of}\mathrm{possible}\mathrm{combinations}-\mathrm{no}.\mathrm{of}\mathrm{groups}\mathrm{with}\mathrm{only}\mathrm{men}$$$$=\left(\begin{array}{c}5+6\\ 4\end{array}\right)-\left(\begin{array}{c}6\\ 4\end{array}\right)$$$$=315$$$$\mathrm{Set}\mathrm{theoretically},\mathrm{if}\mathrm{we}\mathrm{know}\mathrm{the}\mathrm{cardinality}$$$$\mathrm{of}\mathrm{the}\mathrm{finite}\mathrm{universal}\mathrm{set}\mathrm{U}\mathrm{and}\mathrm{the}\mathrm{cardinality}$$$$\mathrm{of}\mathrm{one}\mathrm{of}\mathrm{two}\mathrm{disjoint}\mathrm{sets}\mathrm{A}\mathrm{and}\mathrm{B},\mathrm{and}$$$$\mathrm{U}=\mathrm{A}\cup \mathrm{B},\mathrm{then}\mid \mathrm{A}\mid +\mid \mathrm{B}\mid =\mid \mathrm{U}\mid .$$$$\mathrm{This}\mathrm{situation}\mathrm{has}$$$$\mathrm{U}=\mathrm{all}\mathrm{possible}\mathrm{delegations}$$$$\mathrm{A}=\mathrm{delegations}\mathrm{with}\mathrm{only}\mathrm{men},$$$$\mathrm{B}=\mathrm{delegations}\mathrm{with}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{woman}.$$$$\Rightarrow \mid \mathrm{B}\mid =\mid \mathrm{U}\mid -\mid \mathrm{A}\mid .$$$$$$$$\left(\mathbf{c}\right)\mathrm{W}e\mathrm{can}\mathrm{subtract}\mathrm{from}\mathrm{the}\mathrm{total}\mathrm{number}$$$$\mathrm{of}\mathrm{possible}\mathrm{combinations}\mathrm{those}\mathrm{combinations}$$$$\mathrm{that}\mathrm{are}\mathrm{all}\mathrm{men}\mathrm{and}\mathrm{those}\mathrm{combinations}$$$$\mathrm{that}\mathrm{include}3\mathrm{men}\mathrm{and}1\mathrm{woman}.$$$$\mathrm{By}\mathrm{the}\mathrm{And}\mathrm{counting}\mathrm{principle},$$$$\mathrm{no}.\mathrm{of}\mathrm{combinations}\mathrm{of}3\mathrm{men}\mathrm{and}1\mathrm{woman}$$$$\mathrm{is}\mathrm{equal}\mathrm{to}\left(\begin{array}{c}6\\ 3\end{array}\right)\times \left(\begin{array}{c}5\\ 1\end{array}\right).$$$$\Rightarrow \mathrm{no}.\mathrm{of}\mathrm{ways}\mathrm{required}=\left(\begin{array}{c}5+6\\ 4\end{array}\right)-\{\left(\begin{array}{c}6\\ 4\end{array}\right)+\left(\begin{array}{c}6\\ 3\end{array}\right)\times \left(\begin{array}{c}5\\ 1\end{array}\right)\}$$$$=315-100$$$$=215$$$$\left(\mathbf{d}\right)\mathrm{no}.\mathrm{of}\mathrm{ways}=\left(\begin{array}{c}11-2\\ 3\end{array}\right)+\left(\begin{array}{c}11-2\\ 3\end{array}\right)=2\left(\begin{array}{c}9\\ 3\end{array}\right)=168$$

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