# How many men are not married.

Please help me solve the follow problem: In a group, there are 15 men and 20 women. 17 people are married and 18 people are not married. If the numer of people who are women or married is 27, then how many men are not married. Thanks for your help.

• Did accomplish anything already? – RGS Nov 9 '16 at 13:58
• Have you drawn a diagram? – Arthur Nov 9 '16 at 14:00
• How many people are there? How many are neither women nor married? – Henry Nov 9 '16 at 14:01
• I am NOT inciting a political argument here, but I need to ask for clarification purposes. Are same-sex marriages allowed, or does this problem assume that all marriages are between a man and a woman? What about a man marrying a pencil or some other inanimate object? – teadawg1337 Nov 9 '16 at 14:03
• I have got the example from the book that we work – brain host Nov 9 '16 at 14:03

This is how I'm visualizing it.

Men are represented by the blue circle, women by the red, and married individuals by the green. From here, we can make a system of equations.

There are 15 men $\Rightarrow$ $A + B = 15$.

There are 20 women $\Rightarrow$ $C + D = 20$.

18 are not married $\Rightarrow$ $A + D = 18$.

17 people are married $\Rightarrow$ $B + C = 17$.

27 people are married or women $\Rightarrow$ $B + C + D = 27$.

Solving this sytem of equations will give you all the variables. You are looking for $A$.

• A=8, B=7, C=10, D=10 – brain host Nov 9 '16 at 17:06
• @brainhost You got it. Nice work! – josh Nov 9 '16 at 17:44

Let us say that $G$ is the set of men and $W$ is the set of women. Let $M$ be the set of married people and $NM$ the set of not-married people.

We it is obvious from the context we use the same letter to refer to the number of members of the set.

Then we know:

• $G = 15$

• $W = 20$

• $M = 17$ and hence $NM = 18$

• The set $W \cup M$ has 27 people in it.

You should be able to answer the question with a Venn Diagram and the use of the following identities ($P(X)$ denotes the number of people in the set $X$): (why do they work?)

• $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

• $P(A \cap B) + P(\bar {A} \cap B) = P(B)$

• $\bar{G} = W$

• $\bar{M} = NM$

You are then looking for $P(G \cap NM)$

• Thnx sir, butI do not understand what is the answer – brain host Nov 9 '16 at 16:24
• @brainhost I didn't give you the answer. Weren't you able to figure it out? – RGS Nov 9 '16 at 16:57