Question

There are 85 telephones in the editorial department. How many 2-way connections can be made among the office phones?

Answer 1

have you ever met the quantity \(\binom{n}{k}\) ?

Answer 2

idk. i think i was sleeping in class on this day! haha :/

Answer 3

Let's solve that without this quantity.

We want to count the number of couples (j,k) (j≠k), with j,k=1,...,50.
A direct way to do it is to
1° count the possibilities for the first number
2° count the possibilities for hte second element, given that you can't choose the the first number again.

what is this quantity?

Answer 4

i have no idea. i have been doing math all day i dont think i even know what 4x5 equals anymore.

Answer 5

you don't have to compute the answer, just write the expression.

Answer 6

what expression?! haha, my brain is fried!

Answer 7

for:
1° count the possibilities for the first number
2° count the possibilities for hte second element, given that you can't choose the the first number again.

Answer 8

ok, i will try. haha

Answer 9

(oops i wrote 50 before, but i meant 85)

Answer 10

i think i got it totally wrong. i got 7140. is that wrong?

Answer 11

its' correct.
but it's simpler to write 85 x 84.. (i would never compute that lol)

Answer 12

Now it's not done yet.

Answer 13

omg! i cant believe i got it right! thnx for helping! :)

Answer 14

because, one must 'notice' that choosing , for example, 2, and hten 3, gives the same 'connection' as choosing first 3 and then 2.

Answer 15

huh?

Answer 16

imagine you write down all couples (j,k) as you pass over them all, as we did:
(1° choose one number)
(2° choose another one)

with this technique, you would write (1,2), (1,3), etc..
and later, you will write (2,1), ... but also (3,1)... etc.
But (1,2) and (2,1) represent the same connection. do you see what I mean?

Answer 17

sure. so do yu have to divide the answer by 2? :S

Answer 18

exactly. we counted every connection twice, so the answer is
(85 x 84)/2 . Well done!

Answer 19

ahhh! yay! thnx! :D

Answer 20

this problem is just one example showing the problem of "counting things several times". When you come up with a answer, check if you don't create the same objects several times.. just as we did with the (1,2) and (2,1).

Answer 21

ok! :)

Answer 22

then, if you see you count everything 3 times, just divide the number by 3..
or change the way you count things.
gl :)