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Hint:   a + b = 24 

2222  2228  2232  2238  2252  2258  2282  2288  2292  2298  2322  2328  2332  2338  2352  2358  2382  2388  2392  2398  2522  2528  2532  2538  2552  2558  2582  2588  2592  2598  2822  2828  2832  2838  2852  2858  2882  2888  2892  2898  3222  3228  3232  3238  3252  3258  3282  3288  3292  3298  3322  3328  3332  3338  3352  3358  3382  3388  3392  3398  3522  3528  3532  3538  3552  3558  3582  3588  3592  3598  3822  3828  3832  3838  3852  3858  3882  3888  3892  3898  5222  5228  5232  5238  5252  5258  5282  5288  5292  5298  5322  5328  5332  5338  5352  5358  5382  5388  5392  5398  5522  5528  5532  5538  5552  5558  5582  5588  5592  5598  5822  5828  5832  5838  5852  5858  5882  5888  5892  5898  8222  8228  8232  8238  8252  8258  8282  8288  8292  8298  8322  8328  8332  8338  8352  8358  8382  8388  8392  8398  8522  8528  8532  8538  8552  8558  8582  8588  8592  8598  8822  8828  8832  8838  8852  8858  8882  8888  8892  8898  9222  9228  9232  9238  9252  9258  9282  9288  9292  9298  9322  9328  9332  9338  9352  9358  9382  9388  9392  9398  9522  9528  9532  9538  9552  9558  9582  9588  9592  9598  9822  9828  9832  9838  9852  9858  9882  9888  9892  9898  Total =  200 such numbers

I counted them and got

6 of 1 and 0 of the other      (6,0)  (0,6)

5 of 1 and 1 to the other      (6,2)  (2,6)

4 of 1 and 2 of the other      (6,3)  (3,6)                  also (6,2)  (2,6) again

                                            (6,4)  (4,6)

                                             (5,2)  (2,5)

3 each                                  (5,5)

                                             (6,5)  (5,6)

                                             (3,3)

So I counted    14 possibilities

You can choose them while sitting down, laying down, or standing up. So three ways.

There are 252 numbers between 100 and 999 with AT LEAST one 7 in them.

There are 252 numbers between 100 and 999 with AT LEAST one 9 in them.

5 choices for first digit

    5 choices for 2nd digit

         5 choices for 3rd digit

              5 choices for   4th digit       5 x 5 x 5 x 5 = 5⁴ = 625 numbers

Sorry ....I keep misreading Q's      625 possible numbers    the ones that end in  2 or 8   are even  so, I think 2/5 x 625 = 250 EVEN numbers

The 'x' value of the vertex of the graph of this bowl shaped parabola  will be given by    - b / 2a    where   b = 9   a = -2

   use this value of 'x' in the equation to find the 'y' coordinate of the vertex.....

Jack has 2 more chocolates than Jill.

They can't have very many; so, start low:

If Jack has 3 and Jill has 1, then, when Jill has given one chocolate to Jack, Jack will have 4 and she won't have any, so Jack's number won't be twice as many as Jill.

Try 4 and 2, then 5 and 3, then 6 and 4, then 7 and 5 --

At 7 and 5, if Jack gives one to Jill, they both will have 6 and if Jill gives 1 to Jack, he'll have 8 and she'll have 4 (and Jack will have twice as many as Jill).

Jill has 5.

9!/7!2! = 36

Jack - 1 = Jill +1      (given)    or   Jack = Jill +2     (sub into the next equation)

Jack +1 = 2 (Jill-1)       (given)

Jill +2    +1   = 2 Jill - 2

Jill = 5          Then    Jack =7

Jill still has x chocolates cuz nobody gives candy for free. 

A lot. You probably have to ask them yourself. But be careful of 667. He's pretty obsessive of his possessions.

Yeppers !   c = 3

10 c 3 =   10!/(7!3!) = 120 ways

it is an odd number   because dividing by 2   leaves 1

  it is a multiple of 3   + 2

and an even multiple of 7            7   14   21   28   35    42    49    56    63    70   77    84   91   98    105    112    119   126   133   140  147  

     so for the above rules I have written so far     35   77   119      fit the bill so far

                      divide by 4 to find which leaves 3 remainder :   119

ur basically paying more money for nothing.

1 cuz why would you form more than one executive commitee. 

is c = 3?

The constant c is in the alphabet. I'll list the first six letters for you.

A,B,C,D,E,F

Factor 

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

(x-3)(x-1) (x+5)           - (x +5)(x-1)(x-c) = 0     divide out  (x+5)  and  (x-1)

(x-3)                  -   (x-c )  = 0         think you can take it from here?