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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, 18:42

1

This post received KUDOS

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]

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06 Jan 2010, 11:18

Thanks Atish!
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Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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

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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?

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

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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.

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

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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. 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, 02:37, edited 1 time in total.

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, 09:15

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.

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

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19 Mar 2016, 20:22

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.

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

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

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19 Mar 2016, 21:03

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..!!

gmatclubot

Re: In how many ways can 3-digit numbers be formed selecting 3 d
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19 Mar 2016, 21:03

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