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vipulgoyal: Master | Next Rank: 500 Posts; Posts: 468; Joined: Mon Jul 25, 2011 10:20 pm; Thanked: 29 times; Followed by:4 members

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by vipulgoyal » Tue Mar 12, 2013 11:29 pm

How many 3-digit even numbers can be made using the digits
1, 2, 3, 4, 6, 7, if no digit is repeated?

ans not given
not sure whether GMAT type, but need to review basics

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Source: Beat The GMAT — Problem Solving | Tue Mar 12, 2013 11:29 pm

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Anju@Gurome: GMAT Instructor; Posts: 511; Joined: Wed Aug 11, 2010 9:47 am; Location: Delhi, India; Thanked: 344 times; Followed by:86 members

by Anju@Gurome » Tue Mar 12, 2013 11:58 pm

vipulgoyal wrote:How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

As the integers must be even, the units digit of the integer must be even.
Hence, we have 3 choices (2, 4, 6) for the units digit.

For each of these 3 choices, for other two digits (hundreds and tens), we have 5 and 4 choices.

Hence, total number of such integers = 3*4*5 = 60

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by Anju@Gurome » Wed Mar 13, 2013 12:02 am

Another Method:
Number of 3-digit integers that can be made using the given 6 digits such that no digit is repeated = (Number of ways to select digits for the hundreds place)*(Number of ways to select digits for the tens place)*(Number of ways to select digits for the units place) = (Number of ways to select 1 integer from 6)*(Number of ways to select 1 integer from rest 5)*(Number of ways to select 1 integer from rest 4) = 6*5*4 = 120

As exactly half of the given six digits are even, exactly half of these 120 integers will have even units digit.

Hence, number of even 3-digit integers that can be made using the given 6 digits such that no digit is repeated = 120/2 = 60

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vipulgoyal: Master | Next Rank: 500 Posts; Posts: 468; Joined: Mon Jul 25, 2011 10:20 pm; Thanked: 29 times; Followed by:4 members

by vipulgoyal » Wed Mar 13, 2013 12:12 am

Thanks Anju,Very well explained, I have a query if there would be 5 odds and 3 evens and we have to count no of 3 digit even non repeatable numbers then, woulb be "total 3 digit non repeatable divide by 3/8 " right ans

8*7*6 divided by 3/8

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Anju@Gurome: GMAT Instructor; Posts: 511; Joined: Wed Aug 11, 2010 9:47 am; Location: Delhi, India; Thanked: 344 times; Followed by:86 members

by Anju@Gurome » Wed Mar 13, 2013 12:23 am

vipulgoyal wrote:... if there would be 5 odds and 3 evens and we have to count no of 3 digit even non repeatable numbers then, woulb be "total 3 digit non repeatable divide by 3/8 " right ans

8*7*6 divided by 3/8

Multiplied by NOT divided by.
In the original question the logic is to divide by 6/3, i.e. multiply with 3/6.
Similarly here, multiply with 3/8 or divide by 8/3.

You can cross check the same using the first method also.
We have 3 choices for the units digit and for each of these 3 choices, for other two digits (hundreds and tens), we have 7 and 6 choices.

Hence, total number of such integers = 3*6*7

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vipulgoyal: Master | Next Rank: 500 Posts; Posts: 468; Joined: Mon Jul 25, 2011 10:20 pm; Thanked: 29 times; Followed by:4 members