CPhill

Easy....both EP  and CU  are genuises   !!!!

To get from A to C  we have this possible path

right up up

But we could select the  "ups" to be in any two of the three positions

So  C(3,2)  = 3 possible paths from A to C

To get from  C to B  we have this possible path

right  right right  right    up  up

But we could select the "ups" to occupy any two of the six positions

So C(6,2)  = 15 possible paths from C to B

So...the total possible paths from A to B  = 3 * 15  =  45

Total arrangements -  The number of arrangements where the ARE together

The can be together in  5  ways and for each of these they can be arranged in 2 ways

So

6!  -  2 * 5  =

6!  - 10

720 - 10  =

710  ways

This is the same  as counting the number of  identifiable"words" that can be formed with  four "T's"  and six "F's"

So

[ Total Letters ! ] / [ Repeats of T !    x    Repeats of F  ! ]  =

10!  / [ 4! *  6! [  =  

210 distinct answer keys

Nice, tertre  !!!!!

We want to solve this inequality

Total cost of 1st company < Total cost of second company

.029 N +  4  < .034N      where N is the number of minutes we are looking for

subtract  ,029N from both sides

4 < .005N    divide each side by .005N

4 / .005  < N

800 < N

The first company will be cheaper after 800  minutes

[ Total trees  ! ]   /  [ Number of Goldn De;icious trees! * Number of Bartlett trees ! ]  = 

7! / [ 5! * 2! ]   =   

21 disitnct arrangements

Combos that don't add to  7  are

[ 1. 2 ]  [1,3 ] [1, 4 ] [1,5 ]

[ 2, 3 ]   [ 2, 4 ]

6 ways

If the number ends in 3

We have three choices for the first digit  and two for the second  = 6 numbers

The same logic would apply if the number ended in a "5"

So...   12 

The numer of disitnguishable arrangements  =

4!  / [ 2!  * 2!  ]  =    24 / 4   =    6