Wire problem

[sohailmbaprep]

There are 2 sets ,each having 3 different wires.
Each wire from one set will have to be joined with the one wire from the other set.
In how many ways we can join them . (how many combination possible ?)

[GMATGuruNY]

One set of wires is composed of wires A, B and C. Another set of wires is composed of wires X, Y and Z. A package of wires is to consist of three distinct wire pairs. Each wire pair must be composed of one wire from the first set and one wire from the second set. No type of wire can be used more than once. How many different packages are possible?

I've amended the problem to reflect what I believe it intends to ask.

First pair:
Here, there are 3 options from the first set and 3 options from the second set.
To combine these options, we multiply:
3*3 = 9.

Second pair:
Here, there are 2 remaining options from the first set and 2 remaining options from the second set.
To combine these options, we multiply:
2*2 = 4.

Third pair:
Here, there is 1 remaining option from the first set and 1 remaining option from the second set.
To combine these options, we multiply:
1*1 = 1.

To combine our options for each pair, we multiply the results above:
9*4*1.
Since the ORDER of the pairs doesn't matter -- AX-BY-CZ is the same package as BY-CZ-AX -- we divide by the number of ways to ARRANGE the 3 pairs (3!):
(9*4*1)/(3*2*1) = 6.

For many test-takers, it will be easier simply to write out all of the possible packages:
AX, BY, CZ
AX, BZ, CY
AY, BX, CZ
AY, BZ, CX
AZ, BX, CY
AZ, BY, CX
Total possible packages = 6.

[sohailmbaprep]

There are 2 sets ,each having 3 different wires.
Each wire from one set will have to be joined with the one wire from the other set.
In how many ways we can join them . (how many combination possible ?)

Let the wires in the first set be A, B, and C.
Let the wires in the second set be X, Y, and Z.

The intent of the question stem is not crystal clear.

The most likely interpretation: If one wire must be chosen from each set, how many wire pairs are possible?

Number of options from the first set = 3. (A, B, or C.)
Number of options from the second set = 3. (X, Y or Z.)
To combine these options, we multiply:
3*3 = 9.

For some test-takers, it might be easier simply to write out all of the possible pairs:
AX
AY
AZ
BX
BY
BZ
CX
CY
CZ
Total options = 9.

Another possible interpretation: A package of wires is to consist of three distinct wire pairs. Each wire pair must be composed of one wire from the first set and one wire from the second set. How many different packages are possible?

Here, writing out all of the possible packages might be easiest:
AX, BY, CZ
AX, BZ, CX
AY, BX, CZ
AY, BZ, CX
AZ, BX, CY
AZ, BY, CX
Total options = 6.

I was looking for the second answer thanks
Is there any formula , what if I had 15 wires in each set.

[GMATGuruNY]

Is there any formula , what if I had 15 wires in each set.

Please see my amended post above.