All Answers 358087 Answers

i...dont know about that 46 row thing

Ok!! I am mathtime7 lol. My mom made up the name...

my username is awesomeee for aops

1 + 2 + 3 + ....+ 19 + 20  = n (n + 1) / 2   = (20)(21)/ 2  = 210

What is the 39th number in the row of Pascal's triangle that has 41 numbers?

REALLY?! What's your username?

ok thanks, so... the anwser is 22!

uh...cant anwser that

uh...210. each row has the numbers that the row is for example the 5th row has 5 numbers.

P.S. i do aops too

Hi Melody and Mellie,

For this question  the answer is

$$\Displaystyle \binom{15}{2}+\binom{13}{2}+\binom{11}{2}=238$$

I will explain as soon as I can log in.

Nauseated

depends on what u is. 8 squared is 64

The steps to find the MAD include:

1. find the mean (average)

2. find the difference between each data value and the mean

3. take the absolute value of each difference

4. find the mean (average) of these differences

Source: http://mathbits.com/MathBits/TISection/Statistics1/MAD.html

Sorry, sadly this as well is incorrect. So so so so so so sos sorry!!!!

No, sadly, I don't know the solution and this is incorrect. I'm so sorry for you having to take so much time to solve my problems :(

ok thx lol

Yes!!! This is correct!! Thank you guys sooo sooo much!!

I assume this is :

(1/10)m = (1/11)m + 1/1100       multiply everything through by 1100....this gives....

110m = 100m  + 1       subtract 100m from both sides

10m = 1           divide both sides by 10

m = 1 / 10

How copy? Copy & Paste?

$${\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}} = {\frac{{\mathtt{7}}}{{\mathtt{12}}}} = {\mathtt{0.583\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

This is probably a troll.

But it does take knowledge to know what 1 + 1 is.

I didn't know that at birth.

But some extraordinary people might.

1 + 1=2 really?

Thanks Nauseated,

I am actually wondering if my first answer - which I nutted out without a formula was also correct.

It started with a different premise - that a child could get no lollies, that may be the only reason it was 'wrong'.

Melody, I agree with your answers –both of them.

I’m Nauseated BTW -- The Troll, not my physical condition.

The network connection keeps dropping out, it connects for less than a minute; I can’t login. I’ve tried to post just this over 15 times.

Another for Mellie

In how many ways can we distribute 13 pieces of identical candy to 5 kids, if the two youngest kids are twins and insist on receiving an equal number of pieces?

Ok lets start with a different assumption this time.  Lets assume that every child gets at least one peice.

That means that 3 peices are gone to the elder 3 immediately.  Which means the most the little ones can get is 5 peices each

twins get 5 each      1 way

twins get 4 each, 5 peices left for 3 children     $${\left({\frac{({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)} = {\mathtt{21}}$$    ways

twins get 3 each   7 left for 3 children       $${\left({\frac{({\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}{{\mathtt{7}}{!}{\mathtt{\,\times\,}}({\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{7}}){!}}}\right)} = {\mathtt{36}}$$     ways

twins get   2 each   9 left for the 3 children     $${\left({\frac{({\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}{{\mathtt{9}}{!}{\mathtt{\,\times\,}}({\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{9}}){!}}}\right)} = {\mathtt{55}}$$     ways

twins get 1 each 11 left for the 3 older children       $${\left({\frac{({\mathtt{11}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}){!}}{{\mathtt{11}}{!}{\mathtt{\,\times\,}}({\mathtt{11}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{11}}){!}}}\right)} = {\mathtt{78}}$$

$${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}{\mathtt{\,\small\textbf+\,}}{\mathtt{36}}{\mathtt{\,\small\textbf+\,}}{\mathtt{55}}{\mathtt{\,\small\textbf+\,}}{\mathtt{78}} = {\mathtt{191}}$$   ways  

-----------------------------------------------------------------------------------------------------------------

I still think my original answer might have been correct only it assumed some children might miss out altogether.