Melody

kazza1969,

you need to put different questions on diferent threads :)

Badinage, you are right.     

There are 28 ways for the first, 27 ways for the second and 26 ways for the 3rd   That is 28*27*26

BUT it doesn't matter what order they are chosen.  so you have to divide by 3! = 6

$${\frac{{\mathtt{28}}{\mathtt{\,\times\,}}{\mathtt{27}}{\mathtt{\,\times\,}}{\mathtt{26}}}{{\mathtt{6}}}} = {\mathtt{3\,276}}$$

this is how I would normally do it    

28C3

$${\left({\frac{{\mathtt{28}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{28}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{3\,276}}$$

\\\sqrt9\\

\sqrt{144}\\

\sqrt[3]{64}\\

$$\\\sqrt9\\
\sqrt{144}\\
\sqrt[3]{64}\\$$

There is something else to digest :)

I have added this to our Puzzle Sticky topic   

This really is a fabulous puzzle !!

The area is definitely 49units squared.  I think Fiora was saying the same thing   

It is the diagonal line at the top that causes the deception :)

I used coordinate geometry to find the the y value when x=4 and when x=3  This is the outcome.

You really do lose a litttle 1 unit square in the middle.  

That is because the height has been extended by a skinny rectangle of area 1u^2 is 

ok Mg   

Now put \\ at the front of your first line to get rid of that unwanted indent 

Thank you,

but why didn't you write this on the intial thread   

Thanks Alan and anon - I still trip up on these.    

Thanks MG :)

$${\frac{{{\mathtt{10}}}^{{\mathtt{5}}}}{{\left({{\mathtt{3}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{2}}}\right)}^{{\mathtt{3}}}}}$$

$$\\=\frac{10^5}{3^6\times 10^-6}\\\\
=\frac{10^{5--6}}{3^6}\\\\
=\frac{10^{11}}{3^6}$$