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In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]
 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?
1. 5P3/(2!)
2. 4P3
3. 4^3
4. 4P3+3C1*(3!/2!)
5. 60 × 3!
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Re: P&C - 3-digit numbers..... [#permalink]
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: P&C - 3-digit numbers..... [#permalink]
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: P&C - 3-digit numbers..... [#permalink]
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: P&C - 3-digit numbers..... [#permalink]
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: P&C - 3-digit numbers..... [#permalink]
20 Aug 2009, 13:25
Umm... you want to explain your solution for laypeople? 
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Re: P&C - 3-digit numbers..... [#permalink]
21 Aug 2009, 11:20
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. the SOL looks right.
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Re: P&C - 3-digit numbers..... [#permalink]
22 Aug 2009, 07:10
Can you please exaplain your solution....
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Re: P&C - 3-digit numbers..... [#permalink]
05 Jan 2010, 14:12
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
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Re: P&C - 3-digit numbers..... [#permalink]
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: P&C - 3-digit numbers..... [#permalink]
06 Jan 2010, 12:18
Thanks Atish! 
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Re: P&C - 3-digit numbers..... [#permalink]
06 Jan 2010, 12:28
Thanks for the explanation
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Re: P&C - 3-digit numbers..... [#permalink]
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: P&C - 3-digit numbers..... [#permalink]
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: P&C - 3-digit numbers..... [#permalink]
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: P&C - 3-digit numbers.....
[#permalink]
09 Jan 2010, 09:46
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