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enogh to build a house

(1.5)^-2 = 1 / (1.5)²  = 1/(2.25)  = 4/9

The limits of the 95% inteval are 2 standard deviations above and below the mean.

OK...post them, and we'll see what we can do.....

either that or struggle against the biscuit monster of the treacherous canyons of Haiwaii

you wait until the computer purples

Thanks Bertie, I think I understand perfectly now.  Hopefully I will get it right next time. 

No I really do think I have it.

I would do the second one a little differently.  I'd stick to the 6 possible permutations but I would halve the number of combinations.  It would be easy to miss though. :/

I haven't tried the 2,2,2, on yet - that will be the test LOL  

I see (a) a little differently......

I break this down into 2 tasks.......choose a piece of candy.....choose a luchbox to put it in....

For the first task ...we want to choose one of the 3 candies and one of the 4 lunchboxes to put it in =  C(3,1) * C(4,1)  = 12 ways to do this

So....after we've  chosen the first piece of candy and the first lunchbox....we want to choose one of the 2 remaining pieces of candy and put it into one of the 3 remaining lunchboxes  = C(2,1)*C(3,1)  = 6 ways to do this

And finally, we just want to choose one of the two remaining  lunchboxes to put the last piece of candy into....and there are 2 ways to do this.

So 12 ways x 6 ways x 2 ways = 144 ways

u dont. u dont math.

Technically, the equal sign made it true, but it’s confusing:  Remove errant Binomial from my mind and this post.

Hi Melody

The reason that you get the wrong answer for the 1-1-4 arrangement is that you count some of the possibilities twice.

Suppose that the two odd b***s (if I can call them that !) are labelled A and B, so that there are those two as singles and the others as a group of 4.

They can be placed in the three boxes in 3! = 6 different ways. A-B-4, B-A-4, A-4-B, B-4-A, 4-A-B and 4-B-A.

Now, how many A, B, 4 combinations are there ? The answer is 6C4 = 15, the number of ways in which the group of 4 can be taken from the original 6, (the two singles take care of themselves), not 6*5 = 30.

The problem with the 6*5=30, is that it takes order into account, that is, it sees A,B,4 and B,A,4 as being different and as such covers two of the 6 possibles listed earlier rather than just one.

To look at the 3-3-0 arrangement, suppose that a group of three is chosen and called 3(1), and that the remaining three are called 3(2).

Now suppose that they occupy the first two boxes, so that we have 3(1)-3(2)-0 or 3(2)-3(1)-0. How many possible combinations are there there ? It's tempting to say 6C3 for the first and 6C3 for the second making 2*6C3 = 2*20 = 40 in all. (That, in effect, is what is being done if you say that there are 3!=6 different 3,3,0 arrangements.) That's wrong though, it's actually just 6C3=20, because 6C3 includes both 3(1) and also 3(2).

The 0 can occupy three different positions, so there are 3*20 = 60 combinations in all.

It's easy if you look at it as

box 1    box 2    box 3

3            3           0

3            0           3

0            3           3

There are 6C3 possibilites for each row making 3*6C3 = 60 possibles in all.

The 3!=6, in effect, counts each row twice.

Non-recursive form:

F(n) = 1+(n-1)n(n+1)

F(1) = 1

F(2) = 1+1*2*3 = 7

F(3) = 1 + 2*3*4 = 25

...etc.

. 

I don't know Naus,  I still have a problem with this:

$$\;. \hspace{15pt} \displaystyle \text {c)\; 20 } \hspace{15pt}\leftarrow \hspace{15pt} \left( {\begin{array}{*{20}c} N+k-1 \\ k \\ \end{array}} \right) = \underbrace{\dfrac{(N+k-1)!}{k!}}_{Dist. \ box \ Indist. \ candy }\; \hspace{15pt}| \hspace{15pt} \Text {N=4 \; k=3} \\\\$$

    $$^{(N+k-1)}C_k = \frac{(N+k-1)!}{k!(N+k-1-k)!}=\frac{(N+k-1)!}{k!(N-1)!}$$       

That is still not your formula  :/

A little CDD here. The formula is correct: (4+3-1)!/3! = (6!/3!); the binomial is in error.

Now corrected and noted.

The total cost is given by

$15 (2 * pi * 6 * 15) =

$15 (180 * pi)  = 

about  $8482.30

But Nauseated, you are confusing me.

$$\;. \hspace{15pt} \displaystyle \text {c)\; 20 } \hspace{15pt}\leftarrow \hspace{15pt} \left( {\begin{array}{*{20}c} N \\ k \\ \end{array}} \right) = \underbrace{\dfrac{(N+k-1)!}{k!}}_{Dist. \ box \ Indist. \ candy }\; \hspace{15pt}| \hspace{15pt} \Text {N=4 \; k=3} \\\\$$

$$nCk = 4C3 = \frac{4!}{3!1!}=4$$        

The wood is split into.... 2 + 3 + 4 = 9 equal parts

So....the longest piece is (4/ 9)  of the total length

So (4/9) x 180  = 4 x 20  = 80 cm

logx=6.21781 

In exponential form, we have

10^6.21781  = x   =  about 1,651,239.24

4 years ago, Kate was 4 times younger than her dad. 3 years ago, she was 3 times younger than her dad. Whats her age TODAY? 

Let her age now be= x  .....    and let her father's age now be y  

So

(x - 4)4  = y - 4        →  4x - 16  = y - 4   →   4x - y  = 12       (1)  

(x - 3)3  = y - 3         →  3x - 9 = y - 3      →   3x - y =    6       (2)

Multiplying (2) by -1 and adding it to the first, we have

x = 6 and that's Kate's age now

 And solving for y, we have

4(6) - y = 12

y = 12  and that's her "father's" age now

This doesn't actually make sense......maybe it would make more sense if it were her brother.....but...

The math works out.....

4 years ago, she was 2 and her "father" was  8

3 years ago, she was 3 and her "father" was 9

Hey...it's a strange family.....what can I say????

What was the ambient temperature?

.

 The correct and highly accurate solutions presented below for your entertainment and edification  . . . .

$$\;.\hspace{20pt} \text {a)\; } 4^3 = 64 \; \hspace{15pt}\leftarrow \hspace{15pt} \small \textcolor[rgb]{0,0,0}{\text {No restrictions:}}\\$$

$$\displaystyle \text {b)\; 3} \hspace{15pt}\leftarrow \hspace{15pt} \small \text {Enter "IntegerPartitions [3]" \; into \ Wolfram \ alpha.} \

\hspace{15pt} \text { This will show the values and the symbolic partitions.}\\
\;. \hspace{35pt}\tiny \text { (You may need to use your brain a little more than normal). }\\$$

$$\;. \hspace{15pt} \displaystyle \text {c)\; 20 } \hspace{15pt}\leftarrow \hspace{15pt} \left( {\begin{array}{*{20}c} Removed \ errant \ Binomial \\ \ from \ this \ post \\ \end{array}} \right) = \underbrace{\dfrac{(N+k-1)!}{k!}}_{Dist. \ box \ Indist. \ candy }\; \hspace{15pt}| \hspace{15pt} \Text {N=4 \; k=3} \\\\$$

$$\;. \hspace{15pt} \displaystyle \text {d)\; 7 } \hspace{15pt}\leftarrow \hspace{15pt} \binom {3}{3}\; +\; \binom {3}{2}\; +\; \binom {3}{1}\; =\; 7 \;
\text {(Indist. box; Dist. candy)}\\$$

$$\;. \hspace{15pt}\displaystyle \text {e)\; 40 } \hspace{15pt}\leftarrow \hspace{15pt} \left( \dfrac{4!}{2!}\right)* \left(2 \right) \;+\; \left( \dfrac{4!}{2!}\right) \;+\; \left(4\right)\\$$

$$\text {Reasoning:}\\
\displaystyle \text {All candy in 1 of 4 boxes (4)}\\
\displaystyle \text {2 combinations (XX Y\; \&\; XY X) \;distributed to boxes in (2)(4*3) = (2)(4!/2!) \;=\;(12*2) ways;}\\
\text {and 1 candy to each box distributed to boxes in (1)(4*3*2)/2 = 4!/2! = (12) ways.} \\
\text {Total 4+24+12 = 40 ways.}$$

All this plus or minus CDD . . .

$$\tiny \textcolor[rgb]{1,,0}{\text {(Corrected error in binomial formula for soultion c)}}\\$$

please help i really need this

At constant pressure the volume will be proportional to the absolute temperature.

V2/(273+76) = 1690/(273+43)    (i've assumed your 1.69x103L means 1.69x10³L)

Can you take it from here?

.

@@ End of Day Wrap    Tues 24/3/15   Sydney, Australia    Time 8:15 pm   ♪ ♫

Good evening,

Today our wonderful answerers  were Alan, CPhill, Bertie, Geno3141, MathGod1, Rosala, Jericho, Heureka, TayJay, Gibsonj338, watto, Yoselinedilla, hipbanana and SweatyPerson666.  Thanks everyone. 

The forum was very busy but Chris and I were not answering for much of the day so there are a lot of unanswered question.  I have included a few of the most interesting unaswered ones in the intererest posts. :)

Interest Posts:

This might help:

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