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How many 4 digit even numbers do not use any digit more than

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by Anurag@Gurome » Sat Jun 23, 2012 7:02 am
gmatter2012 wrote:How many 4 digit even numbers do not use any digit more than once
Are you sure of the options?

As we have to form even integer, we have 5 choices for the last digit : 0, 2, 4, 6, and 8

Hence, the last place can be filled in 5 ways.
And, the first place can be filled in 9 ways, the second place can be filled in 8 ways, and the third place can be filled in 7 ways.

Hence, total number of integers = 9*8*7*5 = 2520
Now, there are some integers starting with zero in this lot. We need to discard them.

As the first place has zero, the last place can be filled by 4 ways. And the other two places can be filled by 8 and 7 ways. Hence, total number of integers starting with zero = 1*8*7*4 = 224

Hence, number of required integers = (2520 - 224) = 2296
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by Brent@GMATPrepNow » Sat Jun 23, 2012 7:08 am
gmatter2012 wrote:How many 4 digit even numbers do not use any digit more than once

A. 1720
B. 2160
C. 2240
D. 2460
E. 2520
Take the task and break it into stages, beginning with the most restrictive stage

Stage 1: Select last (units) digit
Stage 2: Select tens digit
Stage 3: Select hundreds digit
Stage 4: Select thousands digit

Stage 1: The digit can be 0, 2, 4, 6, or 8, so there are 5 ways to accomplish stage 1.
Stage 2: Once a digit has been selected, there are 9 digits left to choose from. So there are 9 ways to accomplish stage 2.
Stage 3: 2 digits have been used, so that are now 8 ways to accomplish stage 3.
Stage 4: 3 digits have been used, so that are now 7 ways to accomplish stage 7.
The number of ways to accomplish all 4 stages = 5x9x8x7 = 2520

Important: In the above calculation, we have allowed for the possibility that a zero is selected for the thousands position (stage 4). Since numbers beginning with 0 are not 4-digit numbers, we need to subtract from 2520 all even numbers that begin with a zero. How many such numbers are there? Let's break this task into stages (beginning with the most restrictive stage).

Stage 1: Select 0 as thousands digit.
Stage 2: Select last (units) digit
Stage 3: Select tens digit
Stage 4: Select hundreds digit

Stage 1: Can be accomplished in 1 way
Stage 2: Can be accomplished in 4 ways (2, 4, 6 or 8)
Stage 3: Can be accomplished in 8 ways
Stage 4: Can be accomplished in 7 ways
The number of ways to accomplish all 4 stages = 1x4x8x7 = 224

So, total number of 4-digit even numbers do not use any digit more than once = 2520 - 224 = 2296

Hmmm, this isn't one of the options.

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by Brent@GMATPrepNow » Sat Jun 23, 2012 7:11 am
Please note that my solution is an application of the Fundamental Counting Principle (FCP).
If you're interested, you can watch a free video on the FCP here: https://www.gmatprepnow.com/module/gmat-counting?id=775

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by gmatter2012 » Sat Jun 23, 2012 7:11 am
Please check this method and see if its ok !!

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
Last edited by gmatter2012 on Sat Jun 23, 2012 7:27 am, edited 1 time in total.
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by Brent@GMATPrepNow » Sat Jun 23, 2012 7:13 am
gmatter2012 wrote:Please check this method and see if its ok !!

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 zero }


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
Perfect logic!
I love how these counting question have more than 1 approach.

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by gmatter2012 » Sat Jun 23, 2012 7:37 am
Sorry about the options !! The question was directly copied from the source, the source had the wrong options !!
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by GMATGuruNY » Sat Jun 23, 2012 7:43 am
gmatter2012 wrote:How many 4 digit even numbers do not use any digit more than once
Since it can be helpful to see the different ways a problem can be solved, below is an alternate approach.

Case 1: 0 in the units place
Number of options for the units place = 1. (Must be 0.)
Number of options for the thousands place = 9. (Any digit but 0.)
Number of options for the hundreds place = 8. (Any digit but the two already chosen.)
Number of options for the tens place = 7. (Any digit but the 3 already chosen.)
To combine these options, we multiply:
1*9*8*7 = 504.

Case 2: 2, 4, 6, or 8 in the units place
Number of options for the units place = 4. (Must be 2, 4, 6, or 8.)
Number of options for the thousands place = 8. (Any digit but 0 and the one already chosen.)
Number of options for the hundreds place = 8. (Any digit but the two already chosen.)
Number of options for the tens place = 7. (Any digit but the 3 already chosen.)
To combine these options, we multiply:
4*8*8*7 = 1792.

Total possible integers = 504+1792 = 2296.
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