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How many different 3digit numbers are greater than 299 and do not con [#permalink]
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17 Aug 2017, 01:19
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Re: How many different 3digit numbers are greater than 299 and do not con [#permalink]
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17 Aug 2017, 02:04
Bunuel wrote: How many different 3digit numbers are greater than 299 and do not contain the digits 1, 6, or 8?
(A) 222
(B) 245
(C) 291
(D) 315
(E) 343 3 digits number greater than 299 = 999299 = 700 Three places to be filled to make desired 3 digit number =    Total ways to fill the left most place = 5 choices {3, 4, 5, 7, 9} Total ways to fill the Middle place = 7 choices {All digits except 1, 6 and 8} Total ways to fill the Middle place = 7 choices {All digits except 1, 6 and 8} Total Desired numbers = 5*7*7 = 245 Answer: Option B
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Re: How many different 3digit numbers are greater than 299 and do not con [#permalink]
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17 Aug 2017, 02:12
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Bunuel wrote: How many different 3digit numbers are greater than 299 and do not contain the digits 1, 6, or 8?
(A) 222
(B) 245
(C) 291
(D) 315
(E) 343 Total numbers from 299 to 1000 (excluding both) = 999300+1 = 700 hundred digit = 3,4,5,7 or 9 = 5 options tenth digit = 7 options (0,2,3,4,5,7,9) unit digit = 7 options (0,2,3,4,5,7,9) 5*7*7 = 245 B
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How many different 3digit numbers are greater than 299 and do not con [#permalink]
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17 Aug 2017, 10:50
Bunuel wrote: How many different 3digit numbers are greater than 299 and do not contain the digits 1, 6, or 8?
(A) 222
(B) 245
(C) 291
(D) 315
(E) 343 I didn't calculate total number of threedigit integers between 300 and 1,000. Using slot method to find threedigit numbers that are > 299 and that do not contain 1, 6, and 8 Integers to use if no other restrictions: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9: 7 possibilities if no other restrictions ____ * ____ * ____ FIRST SLOT: 1) Numbers must be greater than 299. First is 300. So first slot's smallest number is 3 2) Not allowed: 0 and 2. Strike them from 7 total above, = 5 possibilities: 3, 4, 5, 7, 9 SECOND SLOT: only first restriction applies. Hence 7 possibilities for second slot THIRD SLOT: same as second slot __ 5__ * __ 7__ * __ 7__ = 245 Answer B
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Re: How many different 3digit numbers are greater than 299 and do not con [#permalink]
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17 Aug 2017, 11:35
Luckisnoexcuse wrote: Bunuel wrote: How many different 3digit numbers are greater than 299 and do not contain the digits 1, 6, or 8?
(A) 222
(B) 245
(C) 291
(D) 315
(E) 343 Total numbers from 299 to 1000 (excluding both) = 999300+1 = 700 hundred digit = 3,4,5,7 or 9 = 5 options tenth digit = 7 options (0,2,3,4,5,7,9) unit digit = 7 options (0,2,3,4,5,7,9) 5*7*7 = 245 B I dont get it what if our first digit (hundenredth digit )is 3 it means now we need (0,2,4,5,7,9) on tenth , WHY DID YOU TOOK 3 for tenth digit? Its MENTIONED WE NEED DIFFERENT 3digit number which means all digit must be different Please help



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Re: How many different 3digit numbers are greater than 299 and do not con [#permalink]
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17 Aug 2017, 12:18
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rocko911 wrote: Luckisnoexcuse wrote: Bunuel wrote: How many different 3digit numbers are greater than 299 and do not contain the digits 1, 6, or 8?
(A) 222
(B) 245
(C) 291
(D) 315
(E) 343 Total numbers from 299 to 1000 (excluding both) = 999300+1 = 700 hundred digit = 3,4,5,7 or 9 = 5 options tenth digit = 7 options (0,2,3,4,5,7,9) unit digit = 7 options (0,2,3,4,5,7,9) 5*7*7 = 245 B I dont get it what if our first digit (hundenredth digit )is 3 it means now we need (0,2,4,5,7,9) on tenth , WHY DID YOU TOOK 3 for tenth digit? Its MENTIONED WE NEED DIFFERENT 3digit number which means all digit must be different Please help Hi rocko911the question says different three digit number and not different digits. for eg. 333 is different from 444, however the digits in both these numbers are same.



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Re: How many different 3digit numbers are greater than 299 and do not con [#permalink]
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04 Dec 2017, 14:02
Hi All, This prompt is essentially just a 'permutation' question, but it does have some minor 'quirks' to it that you have to be careful about to get the correct answer. We're asked for the total number of 3digit numbers that: 1) Can only use the digits 0, 2, 3, 4, 5, 7, 9 2) Must be GREATER than 299. These restrictions will obviously limit the number of possibilities, but there's one other restriction that we have to remember: a 3digit number CANNOT begin with a 0 (e.g. 045 is "45", which is a 2digit number). The first digit can be: 3, 4, 5, 7, 9 (5 options) The second digit can be: 0, 2, 3, 4, 5, 7, 9 (7 options) The third digit can be: 0, 2, 3, 4, 5, 7, 9 (7 options) (5)(7)(7) = 245 total options Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: How many different 3digit numbers are greater than 299 and do not con [#permalink]
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27 Jan 2018, 08:29
GMATinsight wrote: Bunuel wrote: How many different 3digit numbers are greater than 299 and do not contain the digits 1, 6, or 8?
(A) 222
(B) 245
(C) 291
(D) 315
(E) 343 3 digits number greater than 299 = 999299 = 700 Three places to be filled to make desired 3 digit number =    Total ways to fill the left most place = 5 choices {3, 4, 5, 7, 9} Total ways to fill the Middle place = 7 choices {All digits except 1, 6 and 8} Total ways to fill the Middle place = 7 choices {All digits except 1, 6 and 8} Total Desired numbers = 5*7*7 = 245 Answer: Option B I get the above method. But what i initially tried is : for every "hundreds" 3,4,5,7,9. I calculated combinations for "tens" and "units" places for each of them and add them all up. like 3 _ _ is 6C2 which is 30. 30*5=150. What am i doing wrong?



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Re: How many different 3digit numbers are greater than 299 and do not con [#permalink]
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27 Jan 2018, 12:37
Hi Sajjad1093, There are a few errors in your approach: 1) There are actually 7 digits that could be placed in the tens and units 'spots' (not 6). 2) The number 305 and 350 are DIFFERENT results that you have to account for, so this is a permutation (NOT a combination). 3) Digits can be used repeatedly (for example, 333 and 355 are possible options that you have to consider). GMAT assassins aren't born, they're made, Rich
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Re: How many different 3digit numbers are greater than 299 and do not con [#permalink]
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30 Jan 2018, 12:03
EMPOWERgmatRichC wrote: Hi Sajjad1093,
There are a few errors in your approach:
1) There are actually 7 digits that could be placed in the tens and units 'spots' (not 6). 2) The number 305 and 350 are DIFFERENT results that you have to account for, so this is a permutation (NOT a combination). 3) Digits can be used repeatedly (for example, 333 and 355 are possible options that you have to consider).
GMAT assassins aren't born, they're made, Rich my bad i didnt write clearly. so i have assumed the following: 1) I considered "different 3digit numbers" is all digits being different. like 334 or 332 not allowed. 302 and 320 are allowed becuase each of the 3 digit is different. But now you have stated that digits can be repeated, its the figure that needs to be different. 2) So even considering the above correction i have done 7! because considering the limitation on the digits to be used which is 1,6,8 being off limit which leaves us with 7 digits. I have also applied permutations and not combinations. As i have calculated 370 and 307 as unique arrangement Like for 3 _ _ i have calculated 7!/5! because we do not need the rest of 5 digits arrangement as we wont be using them. The above would get me the possible arrangements for "3 hundreds" like 332, 304, 398 etc... which are 42. Thus i now apply the same principle for rest of 4,5,7 and 9 "hundreds" or essentially 42*5=210However considering your 3rd point:When I apply combination on the tens and units places in each hundred, I am not considering 3 4 4 or 3 9 9 arrangement because the digit are only used once in this method and therefore I am falling 35 arrangements short?



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Re: How many different 3digit numbers are greater than 299 and do not con [#permalink]
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30 Jan 2018, 12:25
EMPOWERgmatRichC wrote: Hi Sajjad1093,
There are a few errors in your approach:
1) There are actually 7 digits that could be placed in the tens and units 'spots' (not 6). 2) The number 305 and 350 are DIFFERENT results that you have to account for, so this is a permutation (NOT a combination). 3) Digits can be used repeatedly (for example, 333 and 355 are possible options that you have to consider).
GMAT assassins aren't born, they're made, Rich my bad i didnt write clearly. so i have assumed the following: 1) I considered "different 3digit numbers" is all digits being different. like 334 or 332 not allowed. 302 and 320 are allowed becuase each of the 3 digit is different. But now you have stated that digits can be repeated, its the figure that needs to be different. 2) So even considering the above correction i have done 7! because considering the limitation on the digits to be used which is 1,6,8 being off limit which leaves us with 7 digits. I have also applied permutations and not combinations. As i have calculated 370 and 307 as unique arrangement Like for 3 _ _ i have calculated 7!/5! because we do not need the rest of 5 digits arrangement as we wont be using them. The above would get me the possible arrangements for "3 hundreds" like 332, 304, 398 etc... which are 42. Thus i now apply the same principle for rest of 4,5,7 and 9 "hundreds" or essentially 42*5=210However considering your 3rd point:When I apply combination on the tens and units places in each hundred, I am not considering 3 4 4 or 3 9 9 arrangement because the digit are only used once in this method and therefore I am falling 35 arrangements short?



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Re: How many different 3digit numbers are greater than 299 and do not con [#permalink]
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30 Jan 2018, 13:20
Hi Sajjad1093, In this prompt, digits CAN be used repeatedly, so since ANY of the 7 digits can be in the "ten's spot" and ANY of the 7 digits can be in the "unit's spot", there are (7)(7) options for those two spots. However, your calculation... 7!/5! = (7)(6)... calculates that the two digits CANNOT be repeated. That's why your total is lower than it should be  there are MORE options than you are accounting for with this calculation. GMAT assassins aren't born, they're made, Rich
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