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How many 4 digit even numbers do not use any digit more than
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28 Sep 2010, 10:03
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35% (02:32) correct 65% (02:10) wrong based on 291 sessions
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How many 4 digit even numbers do not use any digit more than once A. 1720 B. 2160 C. 2240 D. 2460 E. 2520
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Re: Counting numbers
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28 Sep 2010, 10:59
rxs0005 wrote: How many 4 digit even numbers do not use any digit more than once
1720
2160
2240
2460
2520 I think OA for this one is wrong. Even 4digit codes (starting with 0 is allowed) with 4 distinct digits {\(abcd\)} = 10*9*8*7/2=2520 (divided by 2 as 4digit codes are half even half odd); Even 4digit codes starting with zero with 4 distinct digits (as we used zero for the first digit so there are total of 101=9 digits left to use) {\(0bcd\)}: {4 choices for \(d\): 2, 4, 6 or 8, zero is out as we used it for the first digit} * {91=8 choices for \(b\)} * {81=7 choices for \(c\)} = 4*8*7 = 224; {Even 4digit codes with 4 distinct digits}  {Even 4digit codes with 4 distinct digits starting with zero} = {Even 4digit numbers with 4 distinct digits} = 2520  224 = 2296.
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Re: Counting numbers
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28 Sep 2010, 12:13
The Way i solved it is
A B C D ( thousand , hundreds, tens, units)
D can be 0 2 4 6 8 ( any of the 5 digits )
A can be anything except (D or 0) so 8 possibilities
C can be anything execpt A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
total ways are 8 * 7 * 8 * 5 = 2240
The only thing different from this logic and bunuel's way is that this has to be a number that means it cannot start with a 0 as the Q is asking 4 digit even "number"



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Re: Counting numbers
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28 Sep 2010, 12:29
rxs0005 wrote: The Way i solved it is
A B C D ( thousand , hundreds, tens, units)
D can be 0 2 4 6 8 ( any of the 5 digits )
A can be anything except (D or 0) so 8 possibilities
C can be anything execpt A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
total ways are 8 * 7 * 8 * 5 = 2240
The only thing different from this logic and bunuel's way is that this has to be a number that means it cannot start with a 0 as the Q is asking 4 digit even "number" The problem with your solution is that when we have 0 for D then the choices for A, B, and C are 9, 8 and 7 respectively and not 8, 8, and 7. So if we do the way you are doing we would have: ABCD: If D is 0, so 1 choice for D, then choices for other letters would be: A9, B8, and C7 > 1*9*8*7=504; If D is 2, 4, 6, or 8, so 4 choices for D, then choices for other letters would be: A8, B8, and C7 > 4*8*8*7=1792; Total: 504+1792=2296.
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Re: Counting numbers
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28 Sep 2010, 10:33
If you allow for 0 in the thousands place, then the answer is easy to find The number of 4digit numbers with no repetition is 10*9*8*7=5040 By symmetry half of these will be even and half odd, hence, number of even numbers is 2520
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Re: Counting numbers
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24 Jun 2012, 03:23
I don't know about others but the very first solution to this problem confused me
Why to start with a zero when it says four digit number , a zero at the beginning will make it a 3 digit no.
here is a more logical approach
4 digit even numbers so first digit cannot be a zero
from 0 to 9 we have 10 digits for each of the four places, please note digits cannot be repeated so each digit has to be distinct.
even numbers will end in a 0 ,2,4,6,8 right?
so lets see if the last digit is zero then
for first digit we have 1 to 9 options{ cannot have 0 here} , second digit we have 8 options , 3 digit we have 7 options and of course units digit i .e has only one option ,0
so 9*8*7*1
same way when last digit is 2
then 8*8*7*1 { 8 options for the thousands place because it cannot be 2 and 0 } { 8 options for the hundredth place because 2 digits have already been reserved } { 7 options for the tens place because 3 digits have already been reserved} { 1 options for the units place because this case is for when unit digit is 2 }
same way when last digit is 4
then 8*8*7*1 { 8 options for the thousands place because it cannot be 4 and 0 } { 8 options for the hundredth place because 2 digits have already been reserved } { 7 options for the tens place because 3 digits have already been reserved} { 1 options for the units place because this case is for when unit digit is 4 }
when last digit is 6
then 8*8*7*1 { 8 options for the thousands place because it cannot be 6 and 0 } { 8 options for the hundredth place because 2 digits have already been reserved } { 7 options for the tens place because 3 digits have already been reserved} { 1 options for the units place because this case is for when unit digit is 6 }
when last digit is 8 then then 8*8*7*1 { 8 options for the thousands place because it cannot be 8 and 0 } { 8 options for the hundredth place because 2 digits have already been reserved } { 7 options for the tens place because 3 digits have already been reserved} { 1 options for the units place because this case is for when unit digit is 8 }
so (8*8*7*1)*4 + 9*8*7*1 = 1792 + 504 = 2296
of course explanation is long but in actual exam all I would do is,
9*8*7*1 + (8*8*7*1)4 = 2296
Hope that helps!



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Re: Counting numbers
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24 Jun 2012, 04:13
shrouded1 wrote: Consider the following cases
Number ending in "0"
Ways to pick number = 1 (fourth digit  is 0) * 9 (third digit  all but 0) * 8 (second digit  all but 0 and third digit) * 7 (first digit  all but 0 and second & third digits)
Number with "0" in tens or hundreds place
Ways to pick number = 4 (fourth digit  {2,4,6,8}) * 2 (either hundreds of tens digit is 0) * 8 (number of ways to pick nonzero of the hundreds/tens digit) * 7 (ways to pick the thousands digit)
Number with no "0"
Ways to pick number = 4 (units digit  no 0 allowed) * 8 (tens digit  one picked and 0 not allowed) * 7 (two picked and 0 not allowed) * 6 (thousands digit  three numbers picked and 0 not allowed)
Total
9*8*7 + 4*2*8*7 + 4*8*7*6 = 56*(9+8+24) = 2296
I suspect this answer is different from the ones you have up there because you are considering numbers starting with 0 as well, which I have not counted I also used the same calculation method...cant figure out my mistake!!!



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Re: Counting numbers
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28 Sep 2010, 10:32
Consider the following cases Number ending in "0"Ways to pick number = 1 (fourth digit  is 0) * 9 (third digit  all but 0) * 8 (second digit  all but 0 and third digit) * 7 (first digit  all but 0 and second & third digits) Number with "0" in tens or hundreds placeWays to pick number = 4 (fourth digit  {2,4,6,8}) * 2 (either hundreds of tens digit is 0) * 8 (number of ways to pick nonzero of the hundreds/tens digit) * 7 (ways to pick the thousands digit) Number with no "0"Ways to pick number = 4 (units digit  no 0 allowed) * 8 (tens digit  one picked and 0 not allowed) * 7 (two picked and 0 not allowed) * 6 (thousands digit  three numbers picked and 0 not allowed) Total9*8*7 + 4*2*8*7 + 4*8*7*6 = 56*(9+8+24) = 2296 I suspect this answer is different from the ones you have up there because you are considering numbers starting with 0 as well, which I have not counted
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Re: Counting numbers
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28 Sep 2010, 13:30
rxs0005 wrote: The Way i solved it is
A B C D ( thousand , hundreds, tens, units)
D can be 0 2 4 6 8 ( any of the 5 digits )
A can be anything except (D or 0) so 8 possibilities
C can be anything execpt A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
total ways are 8 * 7 * 8 * 5 = 2240
The only thing different from this logic and bunuel's way is that this has to be a number that means it cannot start with a 0 as the Q is asking 4 digit even "number" Since your solution does not include the numbers that start with 0 (0BCD), you should be getting the same answer as Bunuel and Shrouded.
I have spend about 15 mins on this, was trying to figure out (1) what were you doing wrong (2) why can't we simply start from the thousandth digit and move to the right. I believe the way Bunuel wrote is a little simpler & quicker, as long as you understand that you need to calculate all 4 digit numbers and then subtract the one's that start with zero (or an alternative that I will provide below). (1) what I think is wrong in your approach (you are getting slightly fewer ways )  A B C D ( thousand , hundreds, tens, units)
D can be 0 2 4 6 8 ( any of the 5 digits ) A can be anything except (D or 0) so 8 possibilities  while doing this you have removed extra possibilities. What happens when D itself is 0 ? so you see, some of the A=0 options have already been removed when you said that A can not be same as D. C can be anything except A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities To compensate for the above, you can do this (this is a correction to your approach or can be an alternative to Bunuel's method)  D = 4 ways (2,4,6,8 only, 0 has been removed) A = 8 ways ( can't be equal to D and 0) B = 8 (can't be equal to A & D) C = 7 (can't be equal to A, B & D) Gives = 1792 (4 digits numbers without repetition, but without the ones that end with 0) D = 1 (can only be equal to 0) A = 9 (anything but D, which also covers cases where A=0) B = 8 (anything but A &D) C = 7 (anything but A, B & D) Gives = 504 (4 digits numbers without repetition that end with 0) Total = 1792 + 504 = 2296 Brunel  (2) why can't we simply start from the thousandth digit and move to the right  is this because we need to take the possibilities for the restrictive digits first, such as the ones digit that can only be even ? Edit  realized after posting that Bunuel had posted a reply.
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Re: Counting numbers
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28 Sep 2010, 13:45
adishail wrote: Brunel  (2) why can't we simply start from the thousandth digit and move to the right  is this because we need to take the possibilities for the restrictive digits first, such as the ones digit that can only be even ? Yes, D determines the # of choices for the rest of the digits.
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Re: Counting numbers
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28 Sep 2010, 22:00
Bunuel wrote: rxs0005 wrote: The Way i solved it is
A B C D ( thousand , hundreds, tens, units)
D can be 0 2 4 6 8 ( any of the 5 digits )
A can be anything except (D or 0) so 8 possibilities
C can be anything execpt A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
total ways are 8 * 7 * 8 * 5 = 2240
The only thing different from this logic and bunuel's way is that this has to be a number that means it cannot start with a 0 as the Q is asking 4 digit even "number" The problem with your solution is that when we have 0 for D then the choices for A, B, and C are 9, 8 and 7 respectively and not 8, 8, and 7. So if we do the way you are doing we would have: ABCD: If D is 0, so 1 choice for D, then choices for other letters would be: A9, B8, and C7 > 1*9*8*7=504; If D is 2, 4, 6, or 8, so 4 choices for D, then choices for other letters would be: A8, B8, and C7 > 4*8*8*7=1792; Total: 504+1792=2296. It's better to divide D in 2 types (0 and others) i had wrong answer cause I did not separate it then I got 5*9*8*7 for D, C ,B, A. Tks Bunuel



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Re: Counting numbers
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29 Sep 2010, 00:29
Somebody please explain me what is wrong woth the below method.
ABCD  4digit #
Choices for A = 9 (0 is excluded) Choices for B = 9 (one digit used for A and 0 is now included) Choices for C = 8 Choices for D = 7
# of 4 digit #s with distinct digits = 9*9*8*7 : note that it would contain equal # of even and odd numbers hence # of EVEN #s with distinct digits = 9*9*8*7 / 2 = 2268



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Re: Counting numbers
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29 Sep 2010, 01:22
muralimba wrote: Somebody please explain me what is wrong woth the below method.
ABCD  4digit #
Choices for A = 9 (0 is excluded) Choices for B = 9 (one digit used for A and 0 is now included) Choices for C = 8 Choices for D = 7
# of 4 digit #s with distinct digits = 9*9*8*7 : note that it would contain equal # of even and odd numbers hence # of EVEN #s with distinct digits = 9*9*8*7 / 2 = 2268 I think that the red part is not correct. Consider another example: there are 90 2digit numbers, out of which 81 have distinct digits (minus 11, 22, 33, 44, 55, 66, 77, 88, 99 total of 9 numbers). Now, if we take your approach then even 2digit numbers with distinct digits should be half of all 2digit numbers with distinct digits  81/2=not an integer. Actual # is 90/2=45 (even 2digit numbers) minus 4 (even 2digt number with same digits: 22, 44, 66, 88) = 41. Hope it's clear.
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Re: Counting numbers
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29 Sep 2010, 10:00
thanks for the explanatory notes bunuel.



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Re: Counting numbers
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30 Sep 2010, 00:41
(10*9*8*7)\2 = 2520



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Re: Counting numbers
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30 Sep 2010, 01:21
prashantbacchewar wrote: (10*9*8*7)\2 = 2520 You can't have zero as the first digit of a 4digit number as in this case it'll become 3digit number, so you don't have 10 choice for the first digit. Please see the solutions of this problem in the previous posts.
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Re: Counting numbers
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24 Jun 2012, 04:16
aks220488 wrote: shrouded1 wrote: Consider the following cases
Number ending in "0"
Ways to pick number = 1 (fourth digit  is 0) * 9 (third digit  all but 0) * 8 (second digit  all but 0 and third digit) * 7 (first digit  all but 0 and second & third digits)
Number with "0" in tens or hundreds place
Ways to pick number = 4 (fourth digit  {2,4,6,8}) * 2 (either hundreds of tens digit is 0) * 8 (number of ways to pick nonzero of the hundreds/tens digit) * 7 (ways to pick the thousands digit)
Number with no "0"
Ways to pick number = 4 (units digit  no 0 allowed) * 8 (tens digit  one picked and 0 not allowed) * 7 (two picked and 0 not allowed) * 6 (thousands digit  three numbers picked and 0 not allowed)
Total
9*8*7 + 4*2*8*7 + 4*8*7*6 = 56*(9+8+24) = 2296
I suspect this answer is different from the ones you have up there because you are considering numbers starting with 0 as well, which I have not counted I also used the same calculation method...cant figure out my mistake!!! 2296 is a correct answer, so you've made no mistake.
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How many 4 digit even numbers do not use any digit more than
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21 Sep 2015, 02:16
We must caclulate here 5 cases for the last digit: 0,2,4,6 and 8 because only in this case is the 4 digit number even. Important points here: 1st digit cannot be 0, and we are looking for 4 ditinct digits ! Case 1 last digit is equal 0: 9 x 8 x 7 x 1 = 504 Case 2 last digit is equal 2: 8 x 8 x 7 x 1 = 448 Case 3 last digit is equal 4: 8 x 8 x 7 x 1 = 448 Case 4 last digit is equal 6: 8 x 8 x 7 x 1 = 448 Case 5 last digit is equal 8: 8 x 8 x 7 x 1 = 448  ∑ = 2296 Actually we have a collision of restrictions only between last and first digit for value “0”, so we must test only 2 Cases: Case 1 last digit is equal 0: 9 x 8 x 7 x 1 = 504 Case 2 last digitany other even value: 8 x 8 x 7 x 4 = 4x448 ∑ = 2296
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How many 4 digit even numbers do not use any digit more than
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