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by arjunshn » Tue Apr 19, 2011 12:13 am
In a certain company the board of directors has been given the responsibility to nominate 3 key positions (one director and two identical positions in the senior management).The board will select 1 of 10 candidates eligible to fill the position of director and 2 of 8 candidates eligible to fill 2 identical positions in the senior management.how many different sets of 3 candidates are there to fill 3 key positions assuming that no candidate is eligible for the post of director as well as senior management nominations?

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by Stuart@KaplanGMAT » Tue Apr 19, 2011 12:54 am
arjunshn wrote:In a certain company the board of directors has been given the responsibility to nominate 3 key positions (one director and two identical positions in the senior management).The board will select 1 of 10 candidates eligible to fill the position of director and 2 of 8 candidates eligible to fill 2 identical positions in the senior management.how many different sets of 3 candidates are there to fill 3 key positions assuming that no candidate is eligible for the post of director as well as senior management nominations?
Hi again!

When we're selecting a subgroup out of a larger group, think combinations. As always, let's start by jotting down the combinations formula:

nCk = n!/k!(n-k)!

in which n is the total number of objects and k is the number in our subgroup.

In this question, we're making two independent selections: one for the director and one for the 2 senior managers. Whenever you make multiple independent selections, you MULTIPLY the individual possibilities.

So, for director we're choosing 1 out of 10 and for senior manager we're choosing 2 out of 8. Therefore, our calculation is:

10C1 * 8C2

= 10 * 8!/2!6!

= 10 * (8*7*6!)/2*6!

= 10 * (8*7)/2

= 10 * 28

= 280

Some good combinations shortcuts to remember:

nC0 = 1
nC1 = n
nCn = 1

(We used the second one to quickly determine that 10C1 = 10.)

As an aside, even if all we could figure out was that there are 10 possibilities for director (you don't have to be a math whiz to reason that if 1 person is going to be chosen out of 10 candidates there are 10 possible choices) and we knew that we had to multiply, we could eliminate any choice that wasn't a multiple of 10. On complex questions with multiple steps, start with the easier ones so you can eliminate answers if you get stuck.
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by arjunshn » Tue Apr 19, 2011 1:05 am
thank you
i was doing it this way
10c1*9c2

i din realise that we have to choose from only the 8 candidates who are eligible.

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by bubbliiiiiiii » Tue Apr 19, 2011 2:30 am
Hi Stuart,

Do we have to bother about the statement 'candidates eligible to fill 2 identical positions in the senior management' in the question?

I mean do we have to divide 8C2 by 2 so that we can eliminate the duplicate outcomes since the positions are identical?
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by Ian Stewart » Tue Apr 19, 2011 9:18 am
bubbliiiiiiii wrote:Hi Stuart,

Do we have to bother about the statement 'candidates eligible to fill 2 identical positions in the senior management' in the question?

I mean do we have to divide 8C2 by 2 so that we can eliminate the duplicate outcomes since the positions are identical?
In the expression 8C2, you are *already* dividing by 2 because the order of the two people selected is irrelevant; 8C2 = (8*7)/(2!). 8C2 *means* "the number of ways to choose 2 people from a group of 8 if the order of the people chosen does not matter". So if you are choosing a committee of 2 people, or a board of 2 directors, or a team of 2 people from a group of 8, then 8C2 is the number of ways you can do that. You don't want to divide by 2 again - it's already taken care of.
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by bubbliiiiiiii » Tue Apr 19, 2011 10:22 am
Thanks Ian for the clarification.
Regards,

Pranay