Permutation and Combination-

[rah_pandey]

Please answer this

There are five wires heading into a building: 2 are for Cable TV and 3 are for phone.if 3 wires were chosen at random, how many different ways could exist so that at least 1 of the wires would be for cable TV?
A. 6
B. 7
C. 8
D. 9
E. 10

it would be nice if you could give working for the problem also.

[sinhavis]

Atleast one wire should be of cable tv and there are maximum 2 wires for cable tv. Thus, it gives us 2 cases...
1) one cable tv wire and 2 phone wires and
2) 2 cable tv wires and 1 phone wire.

Thus for the the first case...
No. of ways will be 2C1(for cable tv)* 3C2(for phone) = 6

for 2nd case...
no. of ways will be 2C2(for cable tv) * 3C1(for phone) = 3

thus total number of ways = 6+3=9

[RonPurewal]

Please answer this

There are five wires heading into a building: 2 are for Cable TV and 3 are for phone.if 3 wires were chosen at random, how many different ways could exist so that at least 1 of the wires would be for cable TV?
A. 6
B. 7
C. 8
D. 9
E. 10

it would be nice if you could give working for the problem also.

notice the signal words, AT LEAST.

"at least" is a complex event - one, or two, or ..., all the way up to the maximum. in some problems, that's a LOT of cases. (not that many in this problem - just one or two - but that's still two cases.

the opposite of "at least one" is NONE, which is a simple event.

so, here's an alternate way to solve this one:

* there are 5! / (3!/2!) ways TOTAL to pick 3 wires.
(if you aren't good at these factorial ditties, you could always do this the "brute force" way, by just making a list.)

* there is EXACTLY ONE way to pick NO wires for cable tv.
(i.e., pick all 3 phone wires)

* so, ten minus one = nine.