bkobilov wrote:Certain company employs 6 senior officers and 4 junior officers. If a committee is to be selected that is made up of 3 senior and 1 junior officer, how many different committees are possible?

a. 8
b. 24
c. 58
d. 80
e. 210

thank you

B: 24

for a problem that asks you arrangements from 2 different groups, do them each individually and add the results.

so for the first one 6!/3!3! = 20
second one 4!/1!3! = 4

therefore the answer is 24.

hi bkolibov/gurpinder:

an official counting problem that sets a classic trap in the final step. gurpinder's work is perfect until the end where he makes the mistake of adding instead of multiplying. this is one of the most common traps that I have seen used by the test makers for these types of combinations.

first determine how many unique combinations of senior officers are possible: 6C3 = 20 (as gurpinder calculated)

second determine how many unique combinations of junior officers are possible: 4C1 = 4 (as gurpinder calculated)

at this point you must multiply to get the correct answer of 80. Why? Remember the counting tree. The only time you add results in counting problems is when you are combining separate counts in problems with wording such as "at least", "at most" or "or" and on some restriction problems where you are adding up allowed cases. Imagine that the 20 possibilities for senior officers are labeled with the numbers 1-20 and the 4 possibilities for junior officers labeled A,B,C,D. Here is a sampling of the 80 possibilities:

1A, 1B, 1C, 1D, 2A, 2B, 2C, 2D, 3A, 3B..........20A, 20B, 20C, and 20D 80 unique groups in total. Hope this helps and let me know if I can clarify futher!

chris@veritasprep wrote:hi bkolibov/gurpinder:

an official counting problem that sets a classic trap in the final step. gurpinder's work is perfect until the end where he makes the mistake of adding instead of multiplying. this is one of the most common traps that I have seen used by the test makers for these types of combinations.

first determine how many unique combinations of senior officers are possible: 6C3 = 20 (as gurpinder calculated)

second determine how many unique combinations of junior officers are possible: 4C1 = 4 (as gurpinder calculated)

at this point you must multiply to get the correct answer of 80. Why? Remember the counting tree. The only time you add results in counting problems is when you are combining separate counts in problems with wording such as "at least", "at most" or "or" and on some restriction problems where you are adding up allowed cases. Imagine that the 20 possibilities for senior officers are labeled with the numbers 1-20 and the 4 possibilities for junior officers labeled A,B,C,D. Here is a sampling of the 80 possibilities:

1A, 1B, 1C, 1D, 2A, 2B, 2C, 2D, 3A, 3B..........20A, 20B, 20C, and 20D 80 unique groups in total. Hope this helps and let me know if I can clarify futher!

Oh my god. I am soo sorry, I was actually doing an OR probability problem on the other hand and mistakenly added the two values here too. Yes the values should be multiplied.

Thanks Chris!

no worries gurpinder - the best way to avoid making mistakes on test day is to burn them into your brain in practice! obviously, you understand the concept and just rushed the problem. Good example of how important attention to detail and focus are on this damn test! It is valuable for everyone to see the traps set in these problems and this is a common and important one....

I await with dread my first mistake posting on here as an "expert". I am sure it will come as my volume of postings increases!!