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

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

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

Answered by Yozzii last updated on 04/May/16

(a) no. of ways= (((5+6)),(4) )= (((11)),(4) )=330 (b) no. of ways=total no. of possible combinations−no. of groups with only men = (((5+6)),(4) )− ((6),(4) ) =315 Set theoretically, if we know the cardinality of the finite universal set U and the cardinality of one of two disjoint sets A and B, and U=A∪B,then ∣A∣+∣B∣=∣U∣. This situation has U=all possible delegations A=delegations with only men, B=delegations with at least one woman. ⇒∣B∣=∣U∣−∣A∣. (c)We can subtract from the total number of possible combinations those combinations that are all men and those combinations that include 3 men and 1 woman. By the And counting principle, no. of combinations of 3 men and 1 woman is equal to ((6),(3) )× ((5),(1) ). ⇒no. of ways required= (((5+6)),(4) )−{ ((6),(4) )+ ((6),(3) )× ((5),(1) )} =315−100 =215 (d)no. of ways= (((11−2)),(3) )+ (((11−2)),(3) )=2 ((9),(3) )=168

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

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