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# M22#16

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Senior Manager
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26 Nov 2008, 20:17
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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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14 Oct 2009, 11:39
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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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26 Nov 2008, 22:15
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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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13 Mar 2010, 17:38
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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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09 May 2011, 06:07
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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

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12 May 2010, 10:44
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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

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19 Mar 2010, 08:32
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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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05 May 2010, 13:17
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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.... 855 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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05 May 2010, 19:49
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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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20 May 2010, 08:00
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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

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12 May 2012, 09:21
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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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13 May 2012, 20:55
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alphabeta1234 wrote:

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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04 May 2014, 12:50
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Here we have to find all different digit number between 710 to 999.
first of all we find the no of digits between 800 to 809
800, 801, 802, 803, 804,805,806, 807, 808, 809 total 8 nos are different
therefore between 800 to 899 = 8*10= 80 is not right, because we 8 number twice.
i.e from 880 to 899 so we have to subtract this no from 80 therefore total 72 nos are different
now we can easily find total different digit no between 700 to 999 as 72*3 =216.
Now subtract 9 from 216 (why 9) because 700 to 709 contains 8 different numbers and in question it is written we have to find three digit no greater than 710. here 1 extra no is 701. This makes a total of 8+1=9)
therefore 216-9=207
hence C is the correct ans
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30 Sep 2009, 12:22
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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09 Dec 2009, 09:49
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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10 May 2011, 08:51

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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15 May 2013, 06:01
How many three-digit integers bigger than 710 are there such that all their digits are different?

Here is another way to look at the same solution:

When 100's digits is 7:
10's digits can be 1 through 9 excluding 7. So 10's digit can be selected in 8 ways.
1's digits can be 0 through 9 excluding 7 and the 10's digit. So 1's digit can be selected in 8 ways.
So, count of ways when 100's digit is 7 = 8 * 8 = 64
But exclude 710, so the resulting count will be 64 - 1 = 63

When 100's digits is 8:
10's digits can be 0 through 9 excluding 8. So 10's digit can be selected in 9 ways.
1's digits can be 0 through 9 excluding 8 and the 10's digit. So 1's digit can be selected in 8 ways.
So, count of ways when 100's digit is 8 = 9 * 8 = 72

When 100's digits is 9:
10's digits can be 0 through 9 excluding 9. So 10's digit can be selected in 9 ways.
1's digits can be 0 through 9 excluding 8 and the 10's digit. So 1's digit can be selected in 8 ways.
So, count of ways when 100's digit is 8 = 9 * 8 = 72

Number of count for three-digit numbers greater than 710 = 63 + 72 + 72 = 207 ways
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29 May 2014, 19:55
I just wanna ask, where did the 707?
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30 May 2014, 01:40
Irawaty wrote:
I just wanna ask, where did the 707?

Do you mean why don't we count 707? If yes, then notice that we need the numbers which are greater than 710 and with distinct digits. 707 does not satisfy any of these conditions.

Hope it's clear.
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# M22#16

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