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M20-23

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How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3
B. 6
C. 10
D. 12
E. 18
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Official Solution:

How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3
B. 6
C. 10
D. 12
E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in \(2(C_{3}^{2}) = 6\) different ways.

Answer: B
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New post 02 Nov 2014, 12:39
I solved using "slot" method. 2*3*1=6

Because the last digit must be 5, there is only one option, therefore, I placed a 1 in this slot. the next slot to the left, therefore, only has 3 possible options (2,3, or 4, since we've already placed the 5, and the numbers cannot repeat). The next slot only has 2 possible options. The product of these is 6.

Is there a simple way to solve this in x!/(y!z!) notation?

Thanks

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JackSparr0w wrote:
I solved using "slot" method. 2*3*1=6

Because the last digit must be 5, there is only one option, therefore, I placed a 1 in this slot. the next slot to the left, therefore, only has 3 possible options (2,3, or 4, since we've already placed the 5, and the numbers cannot repeat). The next slot only has 2 possible options. The product of these is 6.

Is there a simple way to solve this in x!/(y!z!) notation?

Thanks


Your approach is correct.

As for "x!/(y!z!) notation", the solution provided uses 3C2, which is 3!/(2!1!)...
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New post 04 Nov 2014, 17:45
Thanks for the clarification. I'm not familiar with the "C" notation.

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JackSparr0w wrote:
Thanks for the clarification. I'm not familiar with the "C" notation.


nCk is choosing k items out of n distinct items so that the order of the selection does not matter and it equals to \(\frac{n!}{k!(n-k)!}\). For example, choosing 3 out of 5, 5C3, is \(\frac{n!}{k!(n-k)!}=\frac{5!}{3!(5-3)!}=10\): we can choose 10 different 3-item groups out of 5 different items.

Check for more here: math-combinatorics-87345.html

Hope it helps.
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I think this question is good and helpful.

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New post 10 Oct 2015, 07:50
Bunuel, I believe your annotation of your original combination may be incorrect.

You stated 2C3, whereas later you mentioned 3C2. Its a typo I believe, or I could be misinterpreting the C sign, where n is at the top and k is at the bottom.

Thanks

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New post 11 Oct 2015, 05:09
TaiwanTaiwan wrote:
Bunuel, I believe your annotation of your original combination may be incorrect.

You stated 2C3, whereas later you mentioned 3C2. Its a typo I believe, or I could be misinterpreting the C sign, where n is at the top and k is at the bottom.

Thanks


Where do I have 2C3?
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I am wondering why one would use the formula 2 * C2,3? Is that the same as 3!/(3-2)! = 6? (In general, n!/(k-n)!) Thanks!

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New post 14 Oct 2017, 06:03
Bunuel wrote:
tmtsn wrote:
I am wondering why one would use the formula 2 * C2,3? Is that the same as 3!/(3-2)! = 6? (In general, n!/(k-n)!) Thanks!


3C2, \(C^2_3\), \(C^3_2\), all mean \(\frac{3!}{2!(3-2)!}\)


Thanks, but why use the combinations formula instead of permutations formula? Why 2 * 3C2 instead of 3!/(3-2)! which both have the answer 6?

Sorry if I'm being unclear..

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New post 14 Oct 2017, 06:06
tmtsn wrote:
Bunuel wrote:
tmtsn wrote:
I am wondering why one would use the formula 2 * C2,3? Is that the same as 3!/(3-2)! = 6? (In general, n!/(k-n)!) Thanks!


3C2, \(C^2_3\), \(C^3_2\), all mean \(\frac{3!}{2!(3-2)!}\)


Thanks, but why use the combinations formula instead of permutations formula? Why 2 * 3C2 instead of 3!/(3-2)! which both have the answer 6?

Sorry if I'm being unclear..


Yes, we could use \(P^2_3=6\), instead, which is the same as \(2*C^2_3\).
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Re: M20-23 [#permalink]

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New post 21 Oct 2017, 03:21
Bunuel wrote:
Official Solution:

How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3
B. 6
C. 10
D. 12
E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in \(2(C_{3}^{2}) = 6\) different ways.

Answer: B


I got the answer right. But why are you using combination? Since the order matters, shouldn't it be a permutation?

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New post 21 Oct 2017, 03:43
aashishagarwal2 wrote:
Bunuel wrote:
Official Solution:

How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3
B. 6
C. 10
D. 12
E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in \(2(C_{3}^{2}) = 6\) different ways.

Answer: B


I got the answer right. But why are you using combination? Since the order matters, shouldn't it be a permutation?


We multiply 3C2 by 2, which takes care of all arrangements.
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New post 21 Oct 2017, 03:46
Ok, understood. Thanks a ton!

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New post 21 Oct 2017, 10:53
Bunuel wrote:
aashishagarwal2 wrote:
Bunuel wrote:
Official Solution:

How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3
B. 6
C. 10
D. 12
E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in \(2(C_{3}^{2}) = 6\) different ways.

Answer: B


I got the answer right. But why are you using combination? Since the order matters, shouldn't it be a permutation?


We multiply 3C2 by 2, which takes care of all arrangements.


Hi Bunuel, I have a follow-up question to this. If multiplying by 2 takes care of all arrangements, is it like every permutation is equal to double of combination (provided no other constraints on permutation)? Like can we just calculate combination and 2x it?

Thanks for being so kind and and answering all our queries with such patience.

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aashishagarwal2 wrote:
Hi Bunuel, I have a follow-up question to this. If multiplying by 2 takes care of all arrangements, is it like every permutation is equal to double of combination (provided no other constraints on permutation)? Like can we just calculate combination and 2x it?

Thanks for being so kind and and answering all our queries with such patience.


No. We multiply here by 2! (2 factorial) because it's the number of arrangements of 2 distinct objects. If there were n distinct objects it would have been n!.
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Re: M20-23   [#permalink] 21 Oct 2017, 11:09
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