Answer to Question #298604 in Discrete Mathematics for John

Question

250 members of a certain society have voted to elect a new chairman. Each member may vote for either one or two candidates. The candidate elected is the one who polls most votes. Three candidates x, y z stood for election and when the votes were counted, it was found that: - 59 voted for y only, 37 voted for z only - 12 voted for x and y, 14 voted for x and z - 147 voted for either x or y or both x and y but not for z - 102 voted for y or z or both but not for x Required i. Present the information in a Venn diagram. (6 Marks) ii. How many voters did not vote? (4 Marks) iii. How many voters voted for x only? (2 Marks) iv. Who won the elections?

Answer 1

Let $N = 250$ be the total members of the society voted for $X, Y, Z$. Let $n(X), n(Y), n(Z)$ denote the number of candidates voted for $X, Y, Z$ respectively. Then,

$\begin{aligned} N &= n(X \cup Y \cup Z) + n(X \cup Y \cup Z)'\qquad (\text{Voted + Not voted})\\ &= n(X) + n(Y) + n(Z) - n(X \cap Y)-n(X \cap Z)- n(Y \cap Z) \\ &\qquad\qquad\qquad\qquad\qquad+ n(X \cap Y \cap Z)+n(X \cup Y \cup Z)'\\ 250 &= n(X) + n(Y) + n(Z) - n(X \cap Y)-n(X \cap Z) - n(Y \cap Z)\\ &\qquad\qquad\qquad\qquad\qquad + n(X \cup Y \cup Z)' \qquad\qquad\qquad\qquad\qquad(1)\\ &(\text{Since each member may vote for either one or two candidates}\\ &~~~~~~~~~~~~n(X \cap Y \cap Z)=0)\\ \end{aligned}$

$12~ \text{voted for X and Y, ~} 14~ \text{voted for X and Z, i.e.,}\\ \begin{aligned} n(X \cap Y) &=12 \qquad\qquad\quad (2)\\ n(X \cap Z) &=14\qquad\qquad\quad (3)\\\\ \end{aligned}\\ 59~ \text{voted for Y only and~ } 37~ \text{voted for Z only, i.e.,}\\ \begin{aligned} n(Y) - n(X \cap Y) - n(Y \cap Z) &=59 \qquad\qquad\quad (4)\\ n(Z) - n(X \cap Z) - n(Y \cap Z) &=37\qquad\qquad\quad (5)\\\\ \end{aligned}\\ 147~ \text{voted for X or Y or both but not for Z }\\ 102~ \text{voted for Y or Z or both but not for X, i.e.,}\\ \begin{aligned} n(X) + n(Y) - n(X \cap Y) - n(X \cap Z) - n(Y \cap Z) &=147 \qquad\qquad (6)\\ n(Y) + n(Z) - n(Y \cap Z) - n(X \cap Z) - n(X \cap Y) &=102\qquad\qquad(7)\\\\ \end{aligned}\\$

Adding (4) and (5), we get

$n(Y) + n(Z) - n(X \cap Y) - n(X \cap Z) - 2\cdot n(Y \cap Z) = 96 \qquad\quad (8)$

Subtracting (8) from (7), we get $n(Y \cap Z) = 6$.

From (4), (5), (6)

$\begin{aligned} n(Y) &= 59 + n(X\cap Y) + n(Y \cap Z) = 59+12+6=77\\ n(Z) &= 37 + n(X\cap Z) + n(Y \cap Z) = 37+14+6=57\\ n(X) & = 147- n(Y) + n(X \cap Y)+ n(X \cap Z) + n(Y \cap Z)\\ &=147 -77+12+14+6 = 102 \end{aligned}$

i)

ii) Number of candidates who did not vote = $n(X \cup Y \cup Z)'$

From (1),

$\begin{aligned} n(X \cup Y \cup Z)' &= 250 - (n(X) + n(Y) + n(Z) - n(X \cap Y)-n(X \cap Z) - n(Y \cap Z))\\ &= 250-(102+77+57-12-14-6)\\ & = 46 \end{aligned}$

iii) Number of members voted for X only = $n(X) - n(X \cap Y) - n(X \cap Z) = 102-12-14 = 76$

iv) Number of votes for X only = 76,

Number of votes for Y only = 59,

Number of votes for Z only = 37.

Therefore, X won the election.