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In how many ways can 3-digit numbers be formed selecting 3 d

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In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 16 Aug 2009, 03:13
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In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

A. 5P3/(2!)
B. 4P3
C. 4^3
D. 4P3+3C1*(3!/2!)
E. 60 × 3!
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Jul 2015, 08:16, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 19 Aug 2009, 20:39
GODSPEED wrote:
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

1. 5P3/(2!)
2. 4P3
3. 4^3
4. 4P3+3C1*(3!/2!)
5. 60 × 3!


My answer is 4P3. We should discard a double digit. We only have 4 distinctive numbers to choose from.
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 19 Aug 2009, 22:45
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tomirisk wrote:
GODSPEED wrote:
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

1. 5P3/(2!)
2. 4P3
3. 4^3
4. 4P3+3C1*(3!/2!)
5. 60 × 3!


My answer is 4P3. We should discard a double digit. We only have 4 distinctive numbers to choose from.


can't discard duplicated digits, for instance, the number can be 121.

the right one should be 5P3-3(3P2+3P1)=33

answer is C.
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 20 Aug 2009, 03:40
flyingbunny wrote:
tomirisk wrote:
GODSPEED wrote:
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

1. 5P3/(2!)
2. 4P3
3. 4^3
4. 4P3+3C1*(3!/2!)
5. 60 × 3!


My answer is 4P3. We should discard a double digit. We only have 4 distinctive numbers to choose from.


can't discard duplicated digits, for instance, the number can be 121.

the right one should be 5P3-3(3P2+3P1)=33

answer is C.


you are right flyingbunny that we can't discard a double digit. But answer C=4^3=64
and your solution 5P3-3(3P2+3P1)= 34. how can it be C?
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 20 Aug 2009, 04:16
ha, 5P3-3(3P2+3P1)=60-3*9=33
answer is 4. 4P3+3C1*(3!/2!)=33.
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 20 Aug 2009, 12:25
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Umm... you want to explain your solution for laypeople? 8-)
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 05 Jan 2010, 18:42
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For those of you who need a detailed explanation
Lets find the number of 3 digit numbers with only single 1.
i.e. 3 digit numbers from 1,2,3,4
i.e. 4P3 (since order is important)

Next we find numbers with two 1s in it.
the third digit can be chosen in 3C1 ways and the numbers can be arranged in !3 ways, but 2 numbers are the same hence we divide by !2 (Formula used - number of ways of arranging p things which have q things of one kind, r things of another kind and so on is !p/(!q+!r+...))
Finally we have 3C1*(!3/!2))

Add the above two to get the answer D.
Hope this helps
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 06 Jan 2010, 11:18
Thanks Atish! :good
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 06 Jan 2010, 11:28
Thanks for the explanation
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 08 Jan 2010, 04:23
I've a doubt regarding the answer and explanation given.
The question does not mention any constrain on the repetition of digits, so we can have 4 distinct choices for each position of the 3-digited number and hence we have
4^3 = 64 such numbers.
Can any one explain why this can't be the answer?

TIA.
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 09 Jan 2010, 06:13
ashueureka wrote:
I've a doubt regarding the answer and explanation given.
The question does not mention any constrain on the repetition of digits, so we can have 4 distinct choices for each position of the 3-digited number and hence we have
4^3 = 64 such numbers.
Can any one explain why this can't be the answer?

TIA.

Quote:
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

It means you just have 1 digit 2, 1 digit 3, 1 digit 4 and 2 digits 1 to form a number.
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 09 Jan 2010, 08:46
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hi expl of correct ans..
there are four different digits so no of ways 3 digits can be chosen... 4P3...
rest 3 digit nos will have two 1's and a digit from rest 3.... 3!/2!(noof ways when two digits are same)*3C1(3 different digits)... so ans is 4P3+3!/2!*3C1...D
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 26 Nov 2015, 09:02
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post 25 Jan 2016, 06:24
GODSPEED wrote:
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

A. 5P3/(2!)
B. 4P3
C. 4^3
D. 4P3+3C1*(3!/2!)
E. 60 × 3!



This problem is written poorly. It has to state that 1,1,2,3,4 is a set of numbers. So you you can use each element (number) for once. From this perspective solution is easy: There are two cases. Numbers with single 1 and numbers with double 1. Which is 4C3 + 3C1 * 3C2 = 13 (Answer D)

But the problem does not state the set and elements condition and it can be understood that you can use numbers 1,2,3,4 as much as you like and double 1 is just to trick people. From this perspective the answer is 4^3
Re: In how many ways can 3-digit numbers be formed selecting 3 d   [#permalink] 25 Jan 2016, 06:24
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