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# How many 7-digit even numbers less than 3,000,000 can be formed using

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Manager
Joined: 25 Aug 2008
Posts: 149
Location: India
WE 1: 3.75 IT
WE 2: 1.0 IT
How many 7-digit even numbers less than 3,000,000 can be formed using  [#permalink]

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12 Aug 2011, 23:15
1
4
00:00

Difficulty:

85% (hard)

Question Stats:

44% (02:36) correct 56% (02:42) wrong based on 56 sessions

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How many 7-digit even numbers less than 3,000,000 can be formed using the following digits: 1, 2, 2, 3, 5, 5, 6?

A. 180
B. 240
C. 210
D. 270
E. 300

I was able to solve this ques using this techniques.

Case 1: First digit is 1 and last digit is 2 or 6.

(1 X 5 X 4 X 3 X 2 X 1 X 3) / (2 X 2) = 90

Case 2: First digit is 2 and last digit is 2 or 6.

(2 X 5 X 4 X 3 X 2 X 1 X 2) / (2 X 2) = 120

Total permutation = 90 + 120 = 210.

Is there an easier way to do this?

_________________
Cheers,
Varun

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Intern
Status: If I play my cards right, I can work this to my advantage
Joined: 18 Jul 2011
Posts: 14
Location: India
Concentration: Operations, Social Entrepreneurship
GMAT Date: 11-12-2011
WE: Information Technology (Telecommunications)
Re: How many 7-digit even numbers less than 3,000,000 can be formed using  [#permalink]

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13 Aug 2011, 00:04
2
the number is less than 3000000 so we will have the digit on the left most side to be either 1 or 2
i.e your number can be 1 _ _ _ _ _ _ or 2 _ _ _ _ _ _

case 1 > 1 _ _ _ _ _ _ here the units digit can be either 2 or 6

when the units digit is 2 i.e 1 _ _ _ _ _ 2
number of ways this can be done would be 5! / 2! (as 5 is repeated twice) = 60

when the units digit is 6 i.e. 1 _ _ _ _ _ 6, number of ways would be 5!/(2! * 2!) {both 2 and 5 repeat twice} = 30

case 2 > 2 _ _ _ _ _ _ (here units digit may be 2 or 6)
number of ways this can be done would be 5! / (2!) for 2 =60
and 5! / 2! for 6 ... = 60

the explanation looks difficult but this technique helps if you are familiar with the concept of permutations when the numbers/alphabets repeat

hope it is clear
Manager
Joined: 25 Aug 2008
Posts: 149
Location: India
WE 1: 3.75 IT
WE 2: 1.0 IT
Re: How many 7-digit even numbers less than 3,000,000 can be formed using  [#permalink]

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13 Aug 2011, 21:12
fortunetellerz,

Thanks for the explanation but this is almost the same way i solved it. Is there any more simpler and easier way to solve it?
_________________
Cheers,
Varun

If you like my post, give me KUDOS!!
Intern
Joined: 05 Dec 2017
Posts: 7
GMAT 1: 570 Q48 V20
Re: How many 7-digit even numbers less than 3,000,000 can be formed using  [#permalink]

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07 Jan 2019, 20:32
varunmaheshwari wrote:
fortunetellerz,

Thanks for the explanation but this is almost the same way i solved it. Is there any more simpler and easier way to solve it?

I would suggest to split the possible alternatives of 7-digit even numbers less than 3,000,000 and calculate the arrangements. I like to use the Anagram Grids (you can review this method Manhattan GMAT guide 5. Number Properties, Chapter 3. Combinatorics).

First:
1,XXX,XX2 - remaining numbers: 2,3,5,5,6 (Five (numerator) numbers which two of them are repeated (5 and 5) (denominator): 5!/2! = 60)

Second:
1,XXX,XX6 - remaining numbers: 2,2,3,5,5 (Five (numerator) numbers which include two repeated digits (2,2 and 5,5) (denominator): 5!/(2!*2!) = 30)

Third:
2,XXX,XX2 - remaining numbers: 1,3,5,5,6 (Five (numerator) numbers which two of them are repeated (5 and 5) (denominator): 5!/2! = 60)

Fourth:
2,XXX,XX6 - remaining numbers: 1,2,3,5,5 (Five (numerator) numbers which two of them are repeated (5 and 5) (denominator): 5!/2! = 60)

Re: How many 7-digit even numbers less than 3,000,000 can be formed using   [#permalink] 07 Jan 2019, 20:32
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