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Question Stats:
44% (02:14) correct
55% (01:19) wrong based on 97 sessions
How many three-digit integers bigger than 710 are there such that all their digits are different? (A) 98 (B) 202 (C) 207 (D) 209 (E) 212 Source: GMAT Club Tests - hardest GMAT questions Not sure here, but I think the OA is wrong. Thanks.
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ventivish wrote: How many three-digit integers bigger than 710 are there such that all their digits are different?
# 98
# 202
# 207
# 209
# 212
Not sure here, but I think the OA is wrong.
Thanks. 3-digit integers bigger than 710 such that all their digits are different = (3 x 9 x 8) = 216 deduct all those 3-digit integers that are above > 699 but < 711 (do not count 700 and 707 becaue we already ecluded them in 216) = 11-2 = 9 so it is 216 - 9 = 207 intresting
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Can you please let me know how you came up with 3*9*8..? Got the right answer by using diff. approach, however do not understand the explanation that is given, that is the same as yours.
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kt00381n wrote: Can you please let me know how you came up with 3*9*8..? hundreds: 3 possible digits (7,8,9) tens: 9 possible digits (from 0 to 9 minus the one digit already selected for the hundreds) units: 8 possible digits (from 0 to 9 minus the two digits already selected for the hundred and tens) 3*8*9=216. But that's for the 700-999 range. Question asks about the 711-999 range. So you have to substract 701, 702, 703, 704, 705, 706, 708, 709, 710 -> 216-9=207.
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powerka wrote: kt00381n wrote: Can you please let me know how you came up with 3*9*8..? hundreds: 3 possible digits (7,8,9) tens: 9 possible digits (from 0 to 9 minus the one digit already selected for the hundreds) units: 8 possible digits (from 0 to 9 minus the two digits already selected for the hundred and tens) 3*8*9=216. But that's for the 700-999 range. Question asks about the 711-999 range. So you have to substract 701, 702, 703, 704, 705, 706, 708, 709, 710 -> 216-9=207. Thanks Powerka for the explanation.
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The question asks for all different digits. In that case dont we have to subtract occurrences of #s like 777, 722, 855, etc. ?
Thanks!
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another way in 7 series - first digit = 7, second = any digit but 7 and 0 = 8, 3 digit = any digit but 7 and second digit = 8
so total numbers = 8 X 8 = 64 but subtract 1 for 710 so = 64 - 1 = 63.
Now total numbers in 8 and 9 series = 2 X 9X 8 = 144
Total = 144 + 63 = 207
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ventivish wrote: How many three-digit integers bigger than 710 are there such that all their digits are different?
# 98
# 202
# 207
# 209
# 212
Not sure here, but I think the OA is wrong.
Thanks. 100th digit 10 digit one's digit 7,8,9 1 to 9 1 to 8 3 * 9 * 9 = 216 - 9 occurences below 710 = 207 is the answer
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Neels wrote: The question asks for all different digits. In that case dont we have to subtract occurrences of #s like 777, 722, 855, etc. ?
Thanks! I was almost going to ask the same question. Take a look. HTU (3*9*8 ) step1: H =7xx, 8xx, and 9xx...7, 8, and 9 make 3 different hundred's digits step2: T = 0 - 9 (excluding either 7, 8 or 9) => 9 digits left. 77x not possible. Got it? step3: U = 0 - 9 (excluding any digit used as H and that for T) 8 digits left.... 85 5 not possible Subtract numbers ( 9 in all) less than 710 excluding 700 and 707 -already taken care of in steps 1 to 3. Thus the calculation 3*9*8 - 9 = 207. Do i make any sense?
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i did it in a slightly roundabout way....
total numbers in consideration: 999-710= 289
Each hundred set of numbers has 10 in-eligible series of numbers like 77*, 88*, 99* => subtract 30
Every set of 10 numbers (710-719 for eg, has 711, 717) has 2 numbers that are in-eligible (and exclude the set of 10 above):
=> 2 *3 (hundreds series) * 9 => 54 so we can subtract that as well
289-30-54 = 205, but we have to add back the 700, 707 numbers since the range in the question is from 710-999.
this is an alternative way (since i saw the pattern repeating and did it this way)- not elegant at all!
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I will go with option c)207
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I will also go with C. What is source and OA?
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To find number of three digit integers that are bigger than 710: (1) The numbers with digit 7 in hundreds place and digit 1 in tens place: 71  Excluding 7, 1 and 0, there can be seven different digits in the units place. Therefore, count of different numbers = 7 (2) The number with digit 7 in hundreds place and greater than or equal to 2 in in tens place: 7 Excluding 7, 1 and 0, there can be seven different digits in the tens place. Excluding 7 and the digit at tens place, there can be eight different digits in the units place. Therefore, count of different numbers = 7 * 8 = 56 (3) The number with digit 8 or 9 in hundreds place: 8 or 9 There can be two different digits in the hundreds place. Excluding the hundreds digit, there can be nine different digits in the tens place. Excluding the hundreds digit and the tens digit, there can be eight different digits in the tens place. Therefore, count of different numbers = 2 * 9 * 8 = 144 Thus, total count of different number bigger than 710 = 7 + 56 + 144 = 207 Answer is C. Consider awarding me with Kudos if this helps!! Cheers!!! 
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[quote="ventivish"]How many three-digit integers bigger than 710 are there such that all their digits are different? (A) 98 (B) 202 (C) 207 (D) 209 (E) 212 No of different 3 digit no greater than 700 = count of (7,8,9) X {count of (0,1,2,3....7,8,9) - 1} X {count of (0,1,2,3....7,8,9) - 2} =3X9X8=216 Now between 700 to 710 three different digit numbers are 701, 702,703...710 total count 9. So the number of three-digit integers bigger than 710 are there such that all their digits are different = 216 -9 =207 C is the answer.
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0,1,2,3,4,5,6,7,8,9 7,8,9 - 3 ways 2nd digit - 9 ways 3rd digit - 8 ways Total = 72 *3 = 216 700 - 710 < 710 = 11 Subtract 2 from this for 700 and 707 So total = 216 - 9 = 207 Answer - C
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Great post Powerka. Really helped me to see it clearly!
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Option C, good question
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Correct Answer C
3-digit integers bigger than 710 such that all their digits are different = (3 x 9 x 8) = 216
deduct all those 3-digit integers that are above > 699 but < 711 (do not count 700 and 707 becaue we already ecluded them in 216) = 11-2 = 9
so it is 216 - 9 = 207
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My approach was a littler different but should yield the same answer but its not. Can someone please help out.
So I broke the possibilities down.
7 1 __ (one slot) The blank space's numbers can be 1 through 9 (hence 9 numbers) but we must subtract out 7,1 because they have already been picked. Therefore a total of 7 numbers
7 _ _ (two slots) The second slot's number can range from 2 through 9 (hence 8 numbers) but we must subtract a 7, therefore 7 possibilities for the second slot. The third slot can be 6 possibilities because we have one less number that went into the second slot. Hence 7*6=42
_ _ _ (three slots) First slot can be 8 though 9 (2 numbers). The second slot can range from 0 though 9 (10 numbers) but we must subtract the number we placed in the first slot, therefore 9 possible numbers. The third slot can range from 0 though 9 (10 numbers) but we must subtract out the two numbers that we picked for the first and second slot, hence 8 total numbers. Therefore we have 2*9*8=144.
Now we add all the numbers together we get 7+42+144=193. What am I missing here?
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alphabeta1234 wrote: My approach was a littler different but should yield the same answer but its not. Can someone please help out.
7 _ _ (two slots) The second slot's number can range from 2 through 9 (hence 8 numbers) but we must subtract a 7, therefore 7 possibilities for the second slot. The third slot can be 6 possibilities because we have one less number that went into the second slot. Hence 7*6=42
Now we add all the numbers together we get 7+42+144=193. What am I missing here? The thing you are missing for your two slot method is the 2 other possibilities for the ones digit, 1 and 0, which can not be duplicated in the tens or hundreds digit. So instead of 6 possibilities, there are actually 8. So 7*8 = 56, which is 14 more than the 42 you got, which is the difference between your result, (193) and the answer (207).
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