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Make a two-way table as follows:
                                            RED                            WHITE
2-DOOR                               7                                        8

4-DOOR                               3                                      12
TOTALS                               10                                     20

So, it looks like you have 3 cars that are red and have 4 doors.

S(k, i)=Stirling Numbers of the Second Kind. They are computed using "Summation Formula" as explained at the bottom of their table. Here is the Summation formula:

 1/3!*∑[(-1)^i * 3 nCr i * (3 - i)^4, i, 0, 3] =6, where 3= Number of boxes, n=Number of balls.

Note: If you don't understand it, I sympathize with you. Your teacher should explain it to you in detail.

I wonder if someone will answer my question?

I did that, but what does the "S(k,i) mean?

Go online to this site and use the formula in the table at the bottom under "Distinct,  identical"

 Permutations and Combinations > Distributing Balls into Boxes

Thank you, CPhill!

Please make sure you have posted the whole question, some of your question is cut off. 

- Daisy

What question?

Can I have some help please? I need this soon!

(10^9 - 1)!!  =   

1 * 3 * 5 * 7 *  ......* (10^9 - 3) * (10^9 -1)  =

The product of all the odd positive interegers  from 1 to 10^9 -1

This product is extremely large.....WolframAlpha  shows  

≈ 1.58084425021110638757272554931990100564787932540... × 10^4282852759

LOL!!!...we're just weird that way.....!!!!

4 / 353  = x  / 5698     cross-multiply

5698 * 4  = 353 * x

22792  = 353 * x      divide both sides by 353

≈ 64.6 (lbs)   =  x

Performing polynomial divsion...we have....

                ax      +  b

 x^3 - 2  [ ax^4   + bx^3  +   0x     +   3  ]

                ax^4                   -2ax

              _______________________

                            bx^3   + 2ax    +  3

                            bx^3                 -2b

                          ________________

                                       2ax  + (3 + 2b)

We  need to  have ax + b  = 2x  - 3

So

ax  +  b  =  2x  - 3

a =2   b = -3

Thanks!

I don't know how any math could be fun, but I'll take your word for it. Lol

Yeah, they are fun!

- Daisy

Since there is no common base number, we will have to make a common exponent number. The GCF of all of these exponents is \(100\), so lets make all of the exponents \(100\). We have:

a) \({25}^{100}\)

b) \({2}^{3(100)} = {8}^{100}\)

c) \({3}^{4(100)} = {81}^{100}\)

d) \({4}^{2(100)} = {16}^{100}\)

e) \({2}^{6(100)} = {64}^{100}\)

By looking at the base numbers, we can tell which is the least and which is the greatest. The numbers from greatest to least is: \(3^{400} , 2^{600}, 25^{100} , 4^{200} , 2^{300}\)

- Daisy

Thanks, Daisy !!!.....Just wanted to try this one , too....I like these kinds of problems ....

x^2  - y^2  = -12   and    x +  y  = 6   (1)

(x + y) ( x - y)  =  -12

6 ( x - y)  = -12

x - y  =  -2      (2)

Add (1)   and (2)   and we have that   2x  = 4  ⇒   x  = 2

And y =  2 + y  = 6   ⇒  y  = 4

So  (x ,y)  = ( 2, 4)

y = x^2  - 8   ⇒  y^2  = x^4 - 16x^2 + 64   (1)

y^2  = -5x + 44    (2)

Set (1)  = (2)   and solve for x

x^4 - 16x^2 + 64  = -5x + 44

x^4 - 16x^2  + 5x  + 20  = 0

x^2 (x^2 - 16) + 5( x +  4)  = 0

x^2 (x + 4)(x - 4) + 5 (x + 4)  = 0

(x + 4) [ x^2 ( x - 4) + 5 ]   =  0

( x + 4) [ x^3 - 4x^2 + 5 ] = 0

x = -4  is one solution

Also  x = -1  is another solution

To find the remaining polynomial  use synthetic division

-1 [ 1  - 4    0     5]

           -1    5    -5

   ____________

     1   -5    5     0

So...the remaining polynomial is  x^2 - 5x + 5....set to  0

x^2-5x + 5  = 0

x^2 - 5x + 25/4  = -5 + 25/4

(x - 5/2)^2  = 5/4     take both roots

x - 5/2  = ± √5 /2

x = 5/2 + √5/2    or     5/2 - √5/2

When  x  = -4, y= (-4)^2 - 8   = 8

           x = -1 , y = (-1)^2 - 8  = -7

           x = 5/2 + √5/2,   y = (5//2  + √5/2)^2  - 8  =  (1/2)(5√5 - 1)= (1/2)(√125 - 1)

            x = 5/2  - √5/2, y = (5/2 - √5/2)^2  - 8  = (-1/2)(5√5 + 1)  = (-1/2)(√125 + 1)

So....the product of the y coordinates is

( 8 * -7 * (1/2)*(-1/2)(√125 - 1)(√125 + 1 ] =

[ -56 * (-1/4) * (125 - 1) ] =

[ 14 * 124 ]  =

1736

We can make \(x^2 - y^2 = -12\) into \((x-y)(x+y) = -12\) by using “Difference of Squares”. Since we know that \(x+y = 6\), we can substitute that in. Now we have \(6(x-y) = -12\). We can divide \(6\) on both sides, which gives us \(x - y = -2\). We also still have the equation \(x + y = 6\). We can use elimination to get rid of \(x\). We will subtract the second equation from the first. This gives us \(-2y = -8\), so \(y = 4\). We can plug this back into any equation to get \(x\). So \(y = 4\) and \(x = 2\).

- Daisy

Correct! Thanks!

x^2 + y^2  = 14x + 48y     rearrange

x^2 - 14x + y^2 - 48y  = 0    complete the square on x, y

(x^2  - 14x + 49) + (y^2 - 48y + 576)  =  49 + 576

(x - 7)^2  +  ( y - 24)^2  =  625

(x - 7)^2  + (y - 24)^2  = 25^2

This is a circle  with a center at  ( 7, 24)  and a radius of 25

The minimum y value will occur  at  ( 7, 24 - 25)  = ( 7, -1)

So...the  minimum y value  =  -1

The total arrangements possible  = 8!

Now...let's consider the arrangements possible if all three DO sit together [in consecutive seats ]

Considering them as  a single entity....they can be arranged  as 3!  = 6 ways

And they can be seated in any 6 positions  [ positions 1 - 6 ]  = 6 ways

And for each of these arrangements the other people can be seated in 5!  = 120 ways

So...the total arrangements possible if they DON'T sit together   =

Possible arrangements  - Arrangements where they sit together  =

8!  - [ 6 * 6 * 5!]  =

40320 - 36*120  =

40320   - 4320  =

36000  ways

a^2  - b^2   = 8   and  ab  = 2 

So 

(a^2 - b^2) (a^2 - b^2)  =  (8) * (8)

a^4  - 2(ab)^2 + b^4  = 64

a^4 + b^4  = 64 + 2(ab)^2

a^4 + b^4  =  64 + 2(2)^2

a^4 + b^4  = 64 + 8

a^4 + b^4  = 72