CPhill

4. What fraction of all the 10-digit numbers with distinct digits have the property that the sum of every pair of neighboring digits is odd?

Here's my best attempt   :

We either have

Even Odd  Even Odd  Even Odd Even Odd  Even Odd        or

Odd Even  Odd Even Odd Even  Odd Even  Odd Even

In  the first case  we have the following choices for each position....note that the first position cannot be 0, so we only have 4 evens to choose from

4 * 5 * 4 * 4 * 3 * 3 * 2 * 2 * 1 * 1  =    5! * 4 * 4!  =  11520  integers

In the second case we have

5 * 5 * 4 * 5 * 3 * 3 * 2 * 2 * 1 * 1  =    5! * 5!  = 14400 integers 

So    11520 + 14400  = 25920

And we have   10^10  - 10^9    =  9000000000   ten digit integers

So....the fraction is      25920 / 9000000000

Nice job, hectictar   !!!

Don't know..

5.Mort walks each day for exercise and likes to vary his path. He starts each day at position 1 on the circle shown, then walks in a straight line to another numbered point on the circle, and continues, before returning to his starting point. If everyday Mort visits at least one point numbered 2 -- 6 but doesn't visit the same point twice (except for the starting and ending point), how many different possible paths can he take?

Since we can draw  a (straight-line) chord between  any two points on a circle.....the number of possible paths is just al lthe permutations  of the set   { 1,2,3,4,5,6,1 }    with 1  as the first and last elements

So....this  is just   5!  =   120 possible paths

OK....we agree  !!!!

Mmmm....I think it might have to do with the fact that we have " different" chairs

Maybe this is the approach, but I'm not sure

Call the chair   colors    

Red     Blue    Green   Yellow  White 

The boys can  occupy the chairs in the following ways

(R B)  

(B G)

(G Y)

(Y W)

(W R) 

And for each of these    the boys can be seated in two ways   =  5 * 2  = 10 ways

And for each of these ways, the girls can be seated in 3!  = 6 ways

So.....maybe   10 * 3!  =  10 * 6   = 60  is your answer ????

3. Two boys and three girls are going to sit around a table with 5 different chairs. If the two boys want to sit together, in how many possible ways can they be seated?

We can anchor the boys in any two  adjacent chairs.....and they can be seated in 2!  =  2 ways in each of these chairs

And for each of these arrangements, the 3  girls can be seated in 3!  = 6 ways

So....the number of possible seating arrangements  is   2!  * 3! =  2 * 6   =    12 ways

5x^2 + 4  = 3x + 9                  

5x^2 - 3x - 5  = 0       use the Quadratic Formula                  

3 ±√[ 109]

___________   =   x       which implies that

        10

10x  =  3 ±√109          subtract 3 from both sides

10x - 3  =  ±√109       square both sides

So

(10x - 3)^2  = ( ±√109)^2  =

109

We have the following possible equations

(x + 4) + (x - 7)   =  (2x + 1)

2x - 3  = 2x - 1       no solution

(x + 4) + (x - 7)  = -(2x + 1)

2x - 3  = -2x - 1

4x = 2

x = 1/2        doesn't satisfy the original equation

- (x + 4)  + (x - 7)  = 2x + 1           

-11 = 2x + 1

-12 = 2x

x = -6   doesn't satisfy the original equation

(x + 4) - (x - 7) =  2x + 1                     

11 = 2x - 1

12 = 2x

6  = x  doesn't satisfy the original equation

-(x + 4)  - (x - 7)  = 2x + 1

-2x - 11  =  2x + 1

-4x  = -12

x = - 3       dosn't satisfy the original equation

(x + 4)  - (x - 7)  =  - (2x+ 1)                     

11 = -2x - 1

12  = -2x

-6 = x    no

-(x + 4) + (x  - 7)  = -(2x + 1)                       

-11 = -2x - 1

-10  = -2x

x = 5

As this graph shows.....this is the only solution : https://www.desmos.com/calculator/3oejnvli4m

Answered here:   https://web2.0calc.com/questions/two-questions_11