The inclusion-exlusion principle

[nami]

Hey! I am having trouble with a problem that uses the i-e princriple. The problem is:

Among a bank’s 214 customers with checking or savings accounts, 189 have checking accounts, 73 have regular savings accounts, 114 have money market savings accounts, and 69 have both checking and regular accounts. No customer is allowed to have both regular and money market savings accounts.

:idea: Where I am at now:

I let |C| = 189, |R| = 73, |M| = 114, |C n R | = 69, and | C u R U M | = 214

I want to find |C n M|, so I said: |C n M| = |C| + |M| - |R - (C u R U M ) | [This is so I can obtain the value of C u M because I didn't think I could do it directly.]

I am lost at this point. :? Am I on the right track? :?:

Thanx.

 

[stapel]

Are you required to use the Inclusion-Exclusion Principle? Because this is very easy to do with a simple Venn diagram....

Thank you.

Eliz.

 

[pka]

\(\displaystyle \begin{array}{rcl}
\left| {C \cup M \cup R} \right| & = & \left| C \right| + \left| M \right| + \left| R \right| - \left| {C \cap M} \right| - \left| {C \cap R} \right| - \left| {R \cap M} \right| + \left| {C \cap M \cap R} \right|\;\mbox{Inclusion/Exclusion formula} \\
214 & = & 189 + 114 + 73 - \left| {C \cap M} \right| - 69 - 0 - 0. \\
\end{array}\)

 

[soroban]

Hello, nami!

Among a bank’s 214 customers with checking or savings accounts,
189 have checking accounts, 73 have regular savings accounts,
114 have money market savings accounts, and 69 have both checking and regular accounts.
No customer is allowed to have both regular and money market savings accounts.

The question seems to be: How many have Checking and Money Marking Savings accounts?

We are told that: \(\displaystyle \,n(C\,\cup\,R\,\cup\,M)\:=\:214\)
. . \(\displaystyle n(C)\:=\:189,\;n(R)\:=\:73,\;n(M)\:=\:114,\;n(C\,\cap\,R)\:=\:69\)

We also know that: \(\displaystyle \,n(R\,\cap\,M)\:=\:0,\;n(C\,\cap\,R\,\cap\,M)\:=\:0\)


Using this 3-set formula:

\(\displaystyle n(C\,\cup\,R\,\cup\,M) \:=\:n(C)\,+\,n(R)\,+\,n(M)\,-\,n(C\,\cap\,R)\,-\,n(R\,\cap\,M)\,-\,n(C\,\cap\,M)\,+\,n(C\,\cap\,R\,\cap\,M)\)
. - . - .\(\displaystyle \downarrow\) . . . . . . . . \(\displaystyle \downarrow\;\;\;\:\downarrow \;\;\;\;\; \downarrow \;\;\;\;\;\;\;\, \downarrow\;\;\;\;\;\;\;\;\,\downarrow\;\;\;\;\;\;\;\;\;\:\downarrow\;\;\;\;\;\;\;\;\;\;\;\downarrow\)
. . . . \(\displaystyle 214\) . . . \(\displaystyle =\;\;189\;+\;73\;\;+\;114\;\,-\;\;\;\:69\;\;\;\,-\;\;\;\;0\;\;\;\:-\;n(C\,\cap\,M)\;\;+\;\;\;\;\;0\)


Therefore: \(\displaystyle \:n(C\,\cap\,M)\:=\:93\)


(Ha! . . . pka beat me to it.)

 

[nami]

Thanks! I went back and saw where I was going wrong.

I looked at the formula, and there are 3 variables: A, B, and C. What threw me off was that there were 4 numbers in the problem, so I figured that there was something else to it.

I do have another quick question, though. Another part to it asks how many just have checking accounts, but no savings account. Would I follow the formula, but just solve for C, leaving |C n R|, |C n M|, |R n M| as 0? I did it this way, and came up with 27?

Do I at least have this one right? :?: Here's hoping.
:wink:

Thanx so much again!