the digits 3, 4, 5, 6, and 9

This topic has expert replies
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

the digits 3, 4, 5, 6, and 9

by sanju09 » Sat Oct 22, 2011 3:56 am
How many even numbers greater than 40,000 may be formed using the digits 3, 4, 5, 6, and 9 if each digit must be used exactly once in each number?
(A) 36
(B) 48
(C) 64
(D) 96
(E) 112
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Source: — Problem Solving |

Senior | Next Rank: 100 Posts
Posts: 65
Joined: Mon Apr 04, 2011 6:22 am
Thanked: 4 times

by krishnakumar.ks » Sat Oct 22, 2011 4:13 am
First digit can have only 4 numbers (4,5,6 and 9)
Second digit can have any of the numbers from the list except the first digit. Hence it can take 4 numbers
Similarly, third, fourth and fifth digits can take 3,2 and 1 numbers respectively.

Hence total number of numbers that can formed using the digits and greater than 40000 is (4*4*3*2*1)= 96

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Sat Oct 22, 2011 4:17 am
sanju09 wrote:How many even numbers greater than 40,000 may be formed using the digits 3, 4, 5, 6, and 9 if each digit must be used exactly once in each number?
Let us calculate the possible number of integers in each of the following cases,
  • 1. The number starts with 5 or 9
    Leftmost place can be filled with either 5 or 9 --> 2
    Rightmost place can be filled with either 4 or 6 --> 2
    Middle 3 places can be filled with any of the rest three integers --> 3! = 6

    Hence, a total of 2*2*6 = 24 numbers are possible.

    2. The number starts with 4 or 6
    Leftmost place can be filled with either 4 or 6 --> 2
    Rightmost place can be filled with 4 or 6 which was not used in the leftmost place --> 1
    Middle 3 places can be filled with any of the rest three integers --> 3! = 6

    Hence, a total of 2*1*6 = 12 numbers are possible.
Hence, a total of (12 + 24) = 36 numbers are possible.

The correct answer is A.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/

User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

by sanju09 » Sat Oct 22, 2011 4:36 am
The correct answer is [spoiler](A)[/spoiler]. The last digit may only be filled by 4 or 6, thus in 2 ways. This leaves 3 remaining numbers for the first digit, 3 more for the second digit, 2 for the third digit and 1 for the fourth digit. 3 · 3 · 2 · 1 · 2 = [spoiler]36[/spoiler].
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

Junior | Next Rank: 30 Posts
Posts: 29
Joined: Fri Oct 14, 2011 10:15 am
Thanked: 4 times
Followed by:1 members

by zooki » Sat Oct 22, 2011 5:02 am
4 or 6 has to be the last digit for the number to be even. and the integers have to be greater than 40000. i. e. for the 1st digit we have 3 digit after picking 4 or 6 as last digit.

3*_ _ _ *2

for the 2nd digit we have 3, {5/9}, {4/6}

i e. 3*3* _ *_ *2

following the similar manner we have: 3*3*2*1*2= 36