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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]

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16 Aug 2009, 04:13
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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, 09:16, 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]

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19 Aug 2009, 21:39
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]

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19 Aug 2009, 23:45
1
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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

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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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20 Aug 2009, 04:40
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

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]

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20 Aug 2009, 05:16
ha, 5P3-3(3P2+3P1)=60-3*9=33
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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20 Aug 2009, 13:25
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Umm... you want to explain your solution for laypeople?
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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05 Jan 2010, 19:42
1
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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))

Hope this helps
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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06 Jan 2010, 12:18
Thanks Atish!
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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06 Jan 2010, 12:28
Thanks for the explanation
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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08 Jan 2010, 05:23
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]

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09 Jan 2010, 07:13
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]

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09 Jan 2010, 09:46
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]

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In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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25 Jan 2016, 07:24
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. Anyways 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 4P3 + 3C1 * 3C2 (Answer D)

Last edited by leve on 18 Feb 2016, 03:37, edited 1 time in total.
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In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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17 Feb 2016, 20:05
excuse me, but WTF is P? where did u see such notations in official gmat questions??
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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17 Feb 2016, 20:32
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Expert's post
mvictor wrote:
excuse me, but WTF is P? where did u see such notations in official gmat questions??

Hi,
P means permutation and is COUSION, or I should say REAL BROTHER of C, combinations..
only thing is P is very concerned about the order/ sequence, so generally turns out to be not only real brother BUT a BIG BRO too of C ..

so when 5C2 means 5!/3!2!..
5P2 means 5!/3!2! * 2! = 5!/3!, as these two selected can be arranged in 2! ways..
so when you select it is C, and when you arrange, it is P..

Hope it clears some air around the the F of WTF, P
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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19 Mar 2016, 10:15
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!

Had it asked for all the different nos instead of 1,1,2,3,4 the ans wuld be: 5P3/2! = 30
In this solution we hav also halved (divided by 2!) the numbers containing 2,3 and 4.
The numbers containing 2,3,4 are 6.
Add 6/2 =3 back to 30 and we get the answer i.e. 30+3=33.

thanks
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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19 Mar 2016, 21:22
chetan2u wrote:
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

I am having trouble understanding this concept. Can you please explain why we cannot consider these as 5 different digits? How does the repetition of a number have an impact? Really grateful for an explanation.
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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19 Mar 2016, 21:32
Expert's post
1
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MeghaP wrote:
chetan2u wrote:
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

I am having trouble understanding this concept. Can you please explain why we cannot consider these as 5 different digits? How does the repetition of a number have an impact? Really grateful for an explanation.

Hi,

Quote:
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!

say we take these as five different digits..
let these be-
1=a, 1=b, 2=c, 3=d, 4=e..

now we have to choose three digits/letters
so these numbers could be:-
abc=112
bac=112
acd=123
bcd=123 and so on

see here you are taking acd and bcd ; and abc - bac as two different 3-digit numbers, but they are the same..
so to avoid REPETITIONS we take them as 4 different digits

so we consider 5- digits given as 4 different digits to find numbers with all different digits..
example 123,124,234 etc

and then we consider similar digits as two digits and make combinations with remaining digits
example
112, 121, 131,113 and so on

Hope it helps

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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]

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19 Mar 2016, 22:03
chetan2u wrote:
MeghaP wrote:
chetan2u wrote:
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

I am having trouble understanding this concept. Can you please explain why we cannot consider these as 5 different digits? How does the repetition of a number have an impact? Really grateful for an explanation.

Hi,

Quote:
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!

say we take these as five different digits..
let these be-
1=a, 1=b, 2=c, 3=d, 4=e..

now we have to choose three digits/letters
so these numbers could be:-
abc=112
bac=112
acd=123
bcd=123 and so on

see here you are taking acd and bcd ; and abc - bac as two different 3-digit numbers, but they are the same..
so to avoid REPETITIONS we take them as 4 different digits

so we consider 5- digits given as 4 different digits to find numbers with all different digits..
example 123,124,234 etc

and then we consider similar digits as two digits and make combinations with remaining digits
example
112, 121, 131,113 and so on

Hope it helps

It makes sense now. Extremely daft of me to not see that. Thank you so much, much appreciated..!!
Re: In how many ways can 3-digit numbers be formed selecting 3 d   [#permalink] 19 Mar 2016, 22:03
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