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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   Updated on: 27 Jul 2015, 09:16
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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!
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D

Originally posted by GODSPEED on 16 Aug 2009, 04:13.
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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In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post   17 Feb 2016, 20:32
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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!


We can form 3-digit numbers in two ways.
1) All digits different :
there are four different digits so number of ways 3 digits can be chosen... 4P3 or 4C3*3!
2) Two digits similar:
These 3-digit numbers will have two 1s and a digit from rest three - 2, 3, and 4....\( \frac{3!}{2!}\)(number of ways when two digits are same)\(*3C1\)(3 different digits)...

so ans is \(4P3+\frac{3!}{2!}*3C1\)...

D

mvictor wrote:
excuse me, but WTF is P? where did u see such notations in official gmat questions??


Hi,
P means permutation and is COUSIN, 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 also a BIG BRO of C ..

so while 5C2 means \(\frac{5!}{3!2!}\), 5P2 means \(\frac{5!}{3!2!} * 2! = \frac{5!}{3!}\), 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 :) :)

MeghaP wrote:
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,

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 & 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 - 1,1,2,3,4- given as 4 different digits - 1,2,3,4- 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] New post   19 Aug 2009, 21:39
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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, 23:45
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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, 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

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, 05:16
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, 13:25
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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, 19:42
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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, 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] New post   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] New post   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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In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post   Updated on: 18 Feb 2016, 03:37
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)

Originally posted by leve on 25 Jan 2016, 07:24.
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] New post   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] New post   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.
Hence D is the answer.

thanks
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post   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] New post   19 Mar 2016, 21:32
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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] New post   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..!! :)
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post   13 Aug 2017, 22:16
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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!


Given digits : 1,1,2,3,4
There can be 2 cases for forming 3 digit numbers.

Case 1 : When 1 is used only once and we need to form the 3-digit number from digits 1,2,3,4 (Nos. like 123,234, etc.)
No. of such numbers = 4P3

Case 2 : When 1 is used twice and 3rd digit is taken from 2,3,4 (Nos. like 113,141, etc.)
No. of such numbers = 3C1*3!/2! (3C1 for selecting one number out of 3 numbers 2,3,4; 3! means arranging the 3 numbers 1,1 and 3rd selected number and it is divided by 2! as there are 2 1's. Due to which many numbers will be similar.)

Total numbers : 4P3 + 3C1*3!/2!

Answer D
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post   18 Aug 2017, 09:06
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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!



We will separate the 3-digit numbers into two types: Those that contain two 1s and those that contain at most one 1.

1) The numbers that contain two 1s:

These numbers will be of the form _11, 1_1, or 11_, where _ is a digit of 2, 3, or 4. So, there are a total of 9 such numbers.

2) The numbers that contain at most one 1:

These numbers will consist of 3 of the digits 1, 2, 3, and 4. Since the order is important, there are 4P3 such numbers.

Scanning the answer choices, we see that D is the correct answer.

Answer: D
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink] New post   19 Aug 2017, 06:12
I am also getting answer C as question does not mention that repetition is not allowed
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Re: In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]
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