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!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?
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.
