# Computer Probability Question

## Homework Statement

[A computer network consisting of five computers is to be formed by connecting each computer to each of the others by a direct ("point-to-point") network cable.

a) How many network cables are needed?

b) Unfortunately, some of the cables may be faulty ("dead") while others are OK ("alive"). How many different "connectivity patterns" are possible? (E.g., "the cable between computers 1 & 3 is alive, but no others are" is one pattern; "1 & 4, but no others" is a different pattern; "only cable {1, 4} is dead" is a third pattern, etc.)

So I know a) should be 5C2 = 10. But, I am very confused with part b)
I am thinking that to do b) we could think of it as a sum of the number of combinations with increasing number of live cables. For instance if we take 3 computers, then the total no. of combinations possible are (#of ways with 0 live cables) + (# of ways with 1 live cable) + (#no. of ways with 2 live cable) + (# of ways with 3 live cable) = which gives 8.

I could do it for 3 because I manually drew it out, but I am unable to think of equations to come up with these values.
#of ways with 0 live cables = always equal to 1
#of ways with 1 live cable = 5C2 (when there are 5 computers)
#of 1 with with 2 live cables = ?

Am I using the right approach to solve this problem?

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Dick
Homework Helper
Do the three cable case a different way. You have two choices for the state of cable 1, two for the state of cable 2 and two for the state of cable 3. How many total? Answer quickly without splitting into cases on how many are live.

Last edited:
Right! Thank you. Reminds of of binary representation. 5 computers, each two possible states 0 or 1. So 2^5 is all the possible number of states. Now to doing the rest of the parts of the problem. Thanks again :)

Dick