Comtemporary Math Problem?

Comments

• paulr2

  June 2021

  A) You just multiply each number, it's the same as if you had 2 shirts, 2 pairs of pants, and 3 pairs of shoes, you wouldn't say you had (2+2+3)! = 5040, you just multiply it.

  4 * 2 * 5

  8 * 5

  40

  I don't know what it means by principle, maybe you'll know it from what I described in question letter A

  C) I first calculated the chance of taking any one route, based on this you could take the reciprocal and this would be your answer. (1/4 * 1/2 * 1/5, since they're four routes to B, two to C, and 5 to D).

  1/4 * 1/2 * 1/5

  1/8 * 1/5

  1/40

  Reciprocal

  40

  I hope this helps. Have a good day.
• elizabeth-m

  June 2021

  A) Draw a diagram showing highways from B to C via D.

  With each of the 2 highways from B to C there are 5 from C to D

  so there is a total of 2x5 =10 routes from C to D.

  Now include A. There are 4 highways fro A to B and each of these

  has 10 routes from B to D.

  Altogether there are 4x2x5=40 routes from A to D.

  In my day it was called the rs principle, implying you multiply.

  You should first try a simpler version.

  2 roads from A to B, 3 roads from B to C and 2 roads from C to D.

  You can then draw a diagram showing all possible routes from A to D.
• cryptogramcorner

  June 2021

  Actually is it 4 x 2 x 5 = 40

  In this case you are computing the permutations of routes, rather then combinations

  because the order matters

  To see how this works, we can do just the first two legs.

  If you call the 4 routes from A to B

  AB1, AB2, AB3, and AB4

  and the two routes from B to C

  BC1 and BC2

  Then for any particular route from A to B, there will 2 routes to get from B to C

  So for example if you take AB1, then you could finish up with either BC1 or BC2, the

  routes would be

  AB1 - BC1

  AB1 - BC2

  and the same way if you started out with AB2

  AB2 - BC1

  AB2 - BC2

  each starting leg has 2 ways to finish, and since there are 4 starting legs, then are

  4 x 2 complete routes from A to C

  and then if you add the 3rd leg, there are 5 ways to go

  AB1 - BC1 - CD1

  AB1 - BC1 - CD2

  AB1 - BC1 - CD3

  AB1 - BC1 - CD4

  AB1 - BC2 - CD5

  and so on for each of the remaining 7 ways to make the first two legs of the trip.