4 seats and 2 students permutation

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valleeny: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Mon Nov 16, 2009 4:54 pmFollowed by: 1 member; Thanked: 8 times

by valleeny » Sat Dec 12, 2009 4:54 am

How many different ways can 2 students be seated in a row of 4 desks, so that there is always at least one empty desk between the students?

2
3
4
6
12

What's wrong with my method?
Number of ways = Total ways without restrictions - no. of ways where 2 students are next to each other
= 4!/2! - 3!/2!
= 12-3 = 9

wrong answer!

2! being that the position of empty chairs doesn't matter

I know there's a simpler way of simply listing out the possibilities but I prefer to do it the mathematical way

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sunil_snath: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Wed Nov 18, 2009 8:41 am; Thanked: 3 times

by sunil_snath » Sat Dec 12, 2009 6:30 am

total number of ways without restriction is 4C2 which is 6.

Now, if 2 of them sit together, consider them as one and now the number of seat options is 3 for them to sit next to each other (ie if we have 4 seats then options are 1-2, 2-3 or 3-4) so it will be 3C1 = 3

number of ways is 6-3=3

Is this right? I hope this is mathematical enough

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valleeny: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Mon Nov 16, 2009 4:54 pmFollowed by: 1 member; Thanked: 8 times

by valleeny » Sat Dec 12, 2009 7:19 am

Hi thanks. The OA is 6 though.

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thephoenix: Legendary Member; Posts: 1560; Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members; Thanked: 137 times

by thephoenix » Sat Dec 12, 2009 7:55 am

valleeny wrote:How many different ways can 2 students be seated in a row of 4 desks, so that there is always at least one empty desk between the students?

2
3
4
6
12

What's wrong with my method?
Number of ways = Total ways without restrictions - no. of ways where 2 students are next to each other
= 4!/2! - 3!/2!
= 12-3 = 9

wrong answer!

2! being that the position of empty chairs doesn't matter

I know there's a simpler way of simply listing out the possibilities but I prefer to do it the mathematical way

i thing no of ways 2 students are next to each other has to be find out like this take both as single unit hence they can be seated in 3 ways now they can be interchanged among themselves in 2 ways hence total no ways is 3*2=6

reqrd no of ways is 12-6=6

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sunil_snath: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Wed Nov 18, 2009 8:41 am; Thanked: 3 times

by sunil_snath » Sat Dec 12, 2009 7:56 am

oops forgot that 2 students could be flipped around in all the combinations

... so 3*2 = 6.

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Testluv: GMAT Instructor; Posts: 1302; Joined: Mon Oct 19, 2009 2:13 pmFollowed by: 164 members; Location: Toronto; Thanked: 539 times; GMAT Score:800

by Testluv » Sat Dec 12, 2009 10:59 am

valleeny wrote:How many different ways can 2 students be seated in a row of 4 desks, so that there is always at least one empty desk between the students?

2
3
4
6
12

What's wrong with my method?
Number of ways = Total ways without restrictions - no. of ways where 2 students are next to each other
= 4!/2! - 3!/2!
= 12-3 = 9

wrong answer!

2! being that the position of empty chairs doesn't matter

I know there's a simpler way of simply listing out the possibilities but I prefer to do it the mathematical way

The number of restricted cases is 6 and not 3.

Imagine four slots:

- - - -

Let's call the students A and B. We can't have AB in the first two, in the middle two or in the final two slots. That's three restricted cases. But we also can't have BA in the first two, in the middle two or in the final two slots. So, we need to multiply 3 by 2. In other words, to account for order, you needed to multiply by 2.

Similarly, total number of cases is all the ways you can put A and B into any of the two slots: 4C2, but again you have to account for order, so: 4C2 *2.

Therefore,

Total = permitted + restricted

4C2*2 = permitted + 3*2

Permitted = 6.

Kaplan Teacher in Toronto

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maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members; Thanked: 82 times; GMAT Score:720

by maihuna » Sat Dec 12, 2009 11:23 am

Try putting them in four sits one by one:

1. At first seat: Choosen in 1 way, the second may be taken in any two ways(3 or 4th): So two arrangements
2. At second Seat: Only one arrangements
3. At third seat: Only one arrangement as above
4. At last seat: Two arrangement as per 1.

So a total of 2+1+1+2 = 6 ways

Charged up again to beat the beast

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valleeny: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Mon Nov 16, 2009 4:54 pmFollowed by: 1 member; Thanked: 8 times

by valleeny » Sat Dec 12, 2009 7:57 pm

Thanks for all responses.

Testluv, how do you go about thinking of 4C2? 4C2 means "Ther are 4 seats. Choose 2 to be seated by 2 students, the order of the students doesn't matter". But the order does matter so it should be 4P2.

4P2 = 12 as is 4C2 * 2 = 12

Both reasoning are correct? Is there a simpler way to think about this?

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sunil_snath: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Wed Nov 18, 2009 8:41 am; Thanked: 3 times

by sunil_snath » Sun Dec 13, 2009 1:03 am

Hi Valleeny, I think permutation is the right approach on this because it involves ordering as well.

so, 4P2 - 3P1 = 12-6 = 6

So answer will still be the same

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Testluv: GMAT Instructor; Posts: 1302; Joined: Mon Oct 19, 2009 2:13 pmFollowed by: 164 members; Location: Toronto; Thanked: 539 times; GMAT Score:800

by Testluv » Sun Dec 13, 2009 11:46 am

valleeny wrote:Thanks for all responses.

Testluv, how do you go about thinking of 4C2? 4C2 means "Ther are 4 seats. Choose 2 to be seated by 2 students, the order of the students doesn't matter". But the order does matter so it should be 4P2.

4P2 = 12 as is 4C2 * 2 = 12

Both reasoning are correct? Is there a simpler way to think about this?

Oh yes, 4p2 is also correct. I think this is the simplest approach.

Kaplan Teacher in Toronto

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luckyleon: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Thu Nov 12, 2009 5:43 pm

by luckyleon » Sun Dec 13, 2009 5:03 pm

Number of total possiblity - exceptions

4P2 - 3*2 = 6

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