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21 Sep 2020, 17:15
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Hi hennaja,
If you've read through the thread of posts, you'll see that there are a number of different ways that you can answer this question. Depending on which method you prefer, you might find it easiest to treat this question as a Probability question (and calculate ALL the possible ways to get "what we want")….
We have 8 people to choose from; we'll pick 4 - we WANT Andrew in our group, we DON'T WANT Karen in our group.
_ _ _ _
If we want Andrew to be the first person we pick, then the other 3 CAN'T be Karen. We end up with the following math (notice that after picking each person, the total number of people remaining decreases):
(1/8)(6/7)(5/6)(4/5) = 4/56 = 1/14
Andrew COULD be the second, third or fourth person chosen though. Here's the math if he's SECOND and we don't pick Karen in any of the other spots...
(6/8)(1/7)(5/6)(4/5) = 4/56 = 1/14
Notice how the product is the SAME.
If he's third, it would still be 1/14
if he's fourth, it would still be 1/14
With ALL of those possibilities, we end up with 1/14 + 1/14 +1/14 + 1/14 = 4/14 = 2/7
GMAT assassins aren't born, they're made,
Rich
If you've read through the thread of posts, you'll see that there are a number of different ways that you can answer this question. Depending on which method you prefer, you might find it easiest to treat this question as a Probability question (and calculate ALL the possible ways to get "what we want")….
We have 8 people to choose from; we'll pick 4 - we WANT Andrew in our group, we DON'T WANT Karen in our group.
_ _ _ _
If we want Andrew to be the first person we pick, then the other 3 CAN'T be Karen. We end up with the following math (notice that after picking each person, the total number of people remaining decreases):
(1/8)(6/7)(5/6)(4/5) = 4/56 = 1/14
Andrew COULD be the second, third or fourth person chosen though. Here's the math if he's SECOND and we don't pick Karen in any of the other spots...
(6/8)(1/7)(5/6)(4/5) = 4/56 = 1/14
Notice how the product is the SAME.
If he's third, it would still be 1/14
if he's fourth, it would still be 1/14
With ALL of those possibilities, we end up with 1/14 + 1/14 +1/14 + 1/14 = 4/14 = 2/7
GMAT assassins aren't born, they're made,
Rich
11 Oct 2020, 15:26
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Total volunteers: 8
To be selected: 4
Total ways: \(^8{C_4} = 70\)
Condition: Andrew will be always there and Karen will be not.[Andrew being always there we are left with 8 - 1 = 7 volunteers and Karen will never be there, therefore, we have now 7 - 1 = 6 volunteers]
Selecting remaining '3' from '6': \(^6{C_3} = 20\)
Probability: \(\frac{20 }{ 70} = \frac{2}{7}\)
Answer D
To be selected: 4
Total ways: \(^8{C_4} = 70\)
Condition: Andrew will be always there and Karen will be not.[Andrew being always there we are left with 8 - 1 = 7 volunteers and Karen will never be there, therefore, we have now 7 - 1 = 6 volunteers]
Selecting remaining '3' from '6': \(^6{C_3} = 20\)
Probability: \(\frac{20 }{ 70} = \frac{2}{7}\)
Answer D
11 Mar 2021, 08:55
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Reinfrank2011
Hi Bunuel,
Can you explain why it's 4!/3! and not just 4!. I do not understand why the "not karens" who are actually individuals and not identical items do not count determining the number of orders. Also, in case there is a page with theory about this aspect, determining the number of orders in a selection, would be great.
Thank you!
