Combinations questions on the GMAT are considered challenging by many test-takers, but once you know the combinations formula and how to use it, most of the challenge is really just deciphering and visualizing what the word problem is asking. Be sure not to get intimidated when you first read a question like this, and try to work through it step by step.
Problem:
Of the 5 distinguishable wires that lead into an apartment, 2 are for cable television service, and 3 are for telephone service. Using these wires, how many distinct combinations of 3 wires are there such that at least 1 of the wires is for cable television?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Solution:
This is a combinations problem, but with an added twist. Instead of just finding how many groups of three wires we can make, we need to find how many groups of three wires we can make that include at least one cable wire.
Before we worry about the twist, however, we can calculate the number of combinations of three wires we can make, regardless of type of wire. To do this we use the combinations formula, which is n!/(k!(n-k)!). Since we are given five wires and we want to create groups of three, n=5 and k=3, making the formula 5!/(3!(5-3)!). This can be simplified in the following way:
5!(3!2!) =
(5x4x3x2x1)/(3x2x1)(2x1)
Once we cancel out numbers that appear on both the top and bottom of the equation, we get:
(5x4)/2 =
20/2 =
10
Now that we know there are ten different combinations in total, we need to eliminate any that do not include at least cable television wire. As we start with only three telephone wires, the sole combination that would not include any cable television wires is one that includes exactly the three telephone wires. Therefore, only one combination exists, from the original 10, that does not include at least one cable television wire. We eliminate this one outcome from our total, leaving us with 9 possible combinations, which is choice (D).
~Bret Ruber
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