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counting in arrangement

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by eccentric » Tue Aug 26, 2008 2:39 am
I’ve got 10 balls with numbers from 1 to 10 on them. I lost balls 3 and 7. From the balls left, how many ways are there to choose 4 elements in an increasing order?

A 336
B 840
C 70
D 210
E 1680
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Re: counting in arrangement

by sudhir3127 » Tue Aug 26, 2008 3:45 am
eccentric wrote:I’ve got 10 balls with numbers from 1 to 10 on them. I lost balls 3 and 7. From the balls left, how many ways are there to choose 4 elements in an increasing order?

A 336
B 840
C 70
D 210
E 1680
IMO its 70.

8C4..

there are 10 cards but 2 are removed

8 cards are left.

Number of ways of selecting 4 cards from 8 cards ( since all the cards are distinct we arrange them in ascending order )

Thus 8C4 = 70

hope that helps

Do let us know the OA
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by parallel_chase » Tue Aug 26, 2008 3:46 am
I agree with Sudhir's reasoning.

8C4=70
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Confused yet again!!!

by rishi235 » Tue Aug 26, 2008 7:33 am
Hi...
Why don't v take 8P4 here as the arrangement is important here?

Also, on the other hand, if we take 8C4...i.e. v r selecting 4 DISTINCT balls from 8 balls...we do not need to re-arrange the balls in ascending as they r in ascending order...Ok got this but still the arrangement is important right?
I can't understand y v r not taking 8P4 here given that the arrangement is important...
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by somail » Tue Aug 26, 2008 3:44 pm
I agree with rishi, shouldn't this use 8P4 since the question states order matters?
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by nervesofsteel » Tue Aug 26, 2008 7:55 pm
IMO E

reason...

We have 8 balls numbering 1,2,4,5,6,8,9,10

and four places to fill so max number of ways of filling the position are
8*7*6*5 = 1680

Please correct me if im wrong.. P&C are not my strongest bet....
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by dellybean » Wed Aug 27, 2008 4:35 pm
1.2. .4.5.6. .8.9.10

It must be less than E. 1680 is the total number of permutations possible with 8 balls, but the answer is smaller because the balls must all be in ascending order. For example, the first ball cannot be ball 8, 9, or 10 because you would run out of balls before filling all four places, and there is only one possible order when starting with ball #6. This is easier to visualize by looking at the schematic above.

Since only half of the balls can be used as the first ball (1,2,4,5), that knocks out about half of the possible permutations.

(8C4)/2 = 1680/2 = 840

IMO B.

What's the OA?
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by s_kaks » Thu Aug 28, 2008 2:33 am
IMO E

Since order is important here, it has to be 8P4
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by Ian Stewart » Thu Aug 28, 2008 8:40 am
Yes, order is important here, but you need to consider it correctly. There are 8C4 ways of choosing four balls. Once those four balls are chosen, there are 4! ways to put them in order. That is, there are

4!*8C4 (which is equal to 8P4)

ways of both choosing four balls *and* putting them in order. *But* only one of these 4! orders is increasing; the remaining 23 orders we might choose are not increasing. There is 8C4 ways of choosing four balls, and only 1 way to put them in increasing order, so the answer, as pointed out above, is 8C4.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by jeffxujian » Fri Aug 29, 2008 12:25 am
Ian Stewart wrote:Yes, order is important here, but you need to consider it correctly. There are 8C4 ways of choosing four balls. Once those four balls are chosen, there are 4! ways to put them in order. That is, there are

4!*8C4 (which is equal to 8P4)

ways of both choosing four balls *and* putting them in order. *But* only one of these 4! orders is increasing; the remaining 23 orders we might choose are not increasing. There is 8C4 ways of choosing four balls, and only 1 way to put them in increasing order, so the answer, as pointed out above, is 8C4.
Can you please explain why there is only one of those 4! is increasing, whereas the remaining are not increasing.
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