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# An equal number of juniors and seniors are trying out for six spots

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Re: An equal number of juniors and seniors are trying out for six spots [#permalink]
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rajk3433 wrote:
@
Bunuel wrote:
sananoor wrote:
An equal number of juniors and seniors are trying out for six spots on the university debating team. If the team must consist of at least four seniors, then how many different possible debating teams can result if five juniors try out?

A) 50
(B) 55
(C) 75
(D) 100
(E) 250

Five juniors and five seniors (an equal number of juniors and seniors) should form 6-member team so that the team consists of at least four seniors (4, or 5 seniors, all 6 cannot be seniors since there are only 5 seniors).

1. 4 seniors and 2 juniors: $$C^4_5*C^2_5=5*10=50$$

2. 5 seniors and 1 juniors: $$C^5_5*C^1_5=1*5=5$$

Total = 50 + 5 = 55.

Here is a Doubt.

Why not 6 Seniors and 0 Juniors?

Aren't we missing 1 more case 6C6x5C0 = 1?

I am new to the GMAT Club, So I Don't know How to tag @Banuel

Hi,

We have an equal number of juniors and seniors. The problem states that we have 5 juniors, thus we only have 5 seniors. To make a team of 6 using a selection of 5 juniors and 5 seniors, the team must consist of at least 1 junior.

Does this make sense? To make a team of 6 seniors, you would need at least one more senior - the problem states that we only have 5.

I hope this helps -
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Re: An equal number of juniors and seniors are trying out for six spots [#permalink]
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sananoor wrote:
An equal number of juniors and seniors are trying out for six spots on the university debating team. If the team must consist of at least four seniors, then how many different possible debating teams can result if five juniors try out?

A) 50
(B) 55
(C) 75
(D) 100
(E) 250

If the team must have AT LEAST 4 seniors, then we must consider two possible cases:
Case i: The team has 4 seniors and 2 juniors
Case ii: The team has 5 seniors and 1 junior

Case i: The team has 4 seniors and 2 juniors
Since the order in which we select the seniors does not matter, we can use combinations.
STAGE 1: We can select 4 seniors from 5 seniors in 5C4 ways (= 5 ways)
STAGE 2: We can select 2 juniors from 5 juniors in 5C2 ways (= 10 ways)
By the Fundamental Counting Principle (FCP), we can complete both stages in (5)(10) ways = 50 ways

Aside: See the video below to learn how to quickly calculate combinations (like 5C2) in your head

Case ii: The team has 5 seniors and 1 junior
STAGE 1: We can select 5 seniors from 5 seniors in 5C5 ways (= 1 way)
STAGE 2: We can select 1 junior from 5 juniors in 5C1 ways (= 5 ways)
By the Fundamental Counting Principle (FCP), we can complete both stages in (1)(5) ways = 5 ways

TOTAL number of outcomes = 50 + 5 = 55

Cheers,
Brent

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Re: An equal number of juniors and seniors are trying out for six spots [#permalink]
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Re: An equal number of juniors and seniors are trying out for six spots [#permalink]
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