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Re: P&C - 3-digit numbers..... [#permalink]
05 Jan 2010, 13:12

flyingbunny wrote:

ha, 5P3-3(3P2+3P1)=60-3*9=33 answer is 4. 4P3+3C1*(3!/2!)=33.

Can you please explain the solution...

Thanks, JT _________________

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: P&C - 3-digit numbers..... [#permalink]
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 _________________

Re: P&C - 3-digit numbers..... [#permalink]
06 Jan 2010, 11:18

Thanks Atish! _________________

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: P&C - 3-digit numbers..... [#permalink]
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: P&C - 3-digit numbers..... [#permalink]
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.

Re: P&C - 3-digit numbers..... [#permalink]
09 Jan 2010, 08:46

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

gmatclubot

Re: P&C - 3-digit numbers.....
[#permalink]
09 Jan 2010, 08:46