BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

700 level COMBINATORICS problem--Please help

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 209
Joined: Thu Jan 12, 2012 12:59 pm

700 level COMBINATORICS problem--Please help

by dddanny2006 » Sat Dec 21, 2013 12:02 pm
A four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?

a) ¼
b) ½
c) ¾
d) 15/16
e) 1/16

Here's my method to do it.

Digits allowed-0,2,3,5,6,7,8,9

No of possible arrangements=8*8*8*8=2096 -----Order matters since its a code but I dont understand what if the code contains the same numbers?


There are 4 conditions possible
1.All numbers are even =4*4*4*4=256
2.3 numbers are even,one is odd =4*4*4*4=256
3.2 are even,2 are odd=4*4*4*4=256
4.1 is even,3 are odd=4*4*4*4=256

256+256+256+256=1024

Therefore 1024/4096 equals 1/4.

Why am I wrong?
Lets look at another problem below-

How many four digit numbers that do not contain the digits 3 or 6 are there?
a 2401 b 3584 c 4096 d 5040 e 7200


7.8.8.8 = 3584

My approach has got to be correct since most problems that Ive solved using the slot method I have got the right answer.
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Sat Dec 21, 2013 1:57 pm
dddanny2006 wrote:A four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?

a) ¼
b) ½
c) ¾
d) 15/16
e) 1/16
P(at least 1 digit is even) = 1 - P(all 4 digits are odd).

Digit options: 0, 2, 3, 5, 6, 7, 9

P(all 4 digits are odd):
P(1st digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(2nd digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(3rd digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(4th digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
Since selecting 4 odd digits requires that the 1st digit be odd AND the 2nd digit be odd AND the 3rd digit be odd AND the 4th digit be odd, we MULTIPLY the fractions:
4/8 * 4/8 * 4/8 * 4/8 = 1/16.

Thus:
P(at least 1 digit is even) = 1 - 1/16 = 15/16.

The correct answer is D.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Master | Next Rank: 500 Posts
Posts: 209
Joined: Thu Jan 12, 2012 12:59 pm

by dddanny2006 » Sat Dec 21, 2013 1:59 pm
Can you please help me out with the Combinatorics method here.I want to know where I went wrong,as it can help me conceptually too.Please Mitch.
GMATGuruNY wrote:
dddanny2006 wrote:A four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?

a) ¼
b) ½
c) ¾
d) 15/16
e) 1/16
P(at least 1 digit is even) = 1 - P(all 4 digits are odd).

Digit options: 0, 2, 3, 5, 6, 7, 9

P(all 4 digits are odd):
P(1st digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(2nd digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(3rd digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(4th digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
Since selecting 4 odd digits requires that the 1st digit be odd AND the 2nd digit be odd AND the 3rd digit be odd AND the 4th digit be odd, we MULTIPLY the fractions:
4/8 * 4/8 * 4/8 * 4/8 = 1/16.

Thus:
P(at least 1 digit is even) = 1 - 1/16 = 15/16.

The correct answer is D.
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Sat Dec 21, 2013 2:24 pm
dddanny2006 wrote:Can you please help me out with the Combinatorics method here.I want to know where I went wrong,as it can help me
Please see my corrections/notes in red:
dddanny2006 wrote:
Here's my method to do it.

Digits allowed-0,2,3,5,6,7,8,9

No of possible arrangements=8*8*8*8=4096

There are 4 conditions possible

1. All numbers are even =4*4*4*4=256

2. 3 numbers are even,one is odd =4*4*4*4=256
There are FOUR ways Case 2 could happen:
The 1st digit could be odd.
The 2nd digit could be odd.
The 3rd digit could be odd.
The 4th digit could be odd.
Thus, the result above must be multiplied by 4:
4*256.

3. 2 are even,2 are odd=4*4*4*4=256
There are SIX ways Case 3 could happen.
The following pairs of digits could be even:
1st and 2nd
1st and 3rd
1st and 4th
2nd and 3rd
2nd and 4th
3rd and 4th.
Thus, the result above must be multiplied by 6:
6*256.

4.1 is even,3 are odd=4*4*4*4=256
There are FOUR ways Case 4 could happen:
The 1st digit could be even.
The 2nd digit could be even.
The 3rd digit could be even.
The 4th digit could be even.
Thus, the result above must be multiplied by 4:
4*256.

Total good cases = 256 + (4*256) + (6*256) + (4*256) = 15*256.

P(good outcome) = (15*256)/4096 = 15/16.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Master | Next Rank: 500 Posts
Posts: 209
Joined: Thu Jan 12, 2012 12:59 pm

by dddanny2006 » Sat Dec 21, 2013 2:28 pm
Thanks Mitch.Lets consider case 2 of the problem where we have 3even and one odd.

I had it as a 4*4*4*4 ---what would this have given me? I was of the opinion that when order doest matter we divide it by the number of interchangeable slots.But here since order is important I left it alone thinking that it would give me all the possible arrangements,but No,we have to firther multiply it by 4.Im a little confused here.Since I treated it as a permutation I should have got all possible arrangements. :(

Also when deciding the number of possible arrangements ie.8*8*8*8,why dont we do the samething thats in red?Here too we dont know if the order matters because it could be that all the 4 numbers could be the same?Right??Why is that a permutation?

For example in the problem below-
How many four digit numbers that do not contain the digits 3 or 6 are there?
a 2401 b 3584 c 4096 d 5040 e 7200


7.8.8.8 = 3584

Here too the number positions could get interchanged right.

We could have 4598 at the same time we may also have a 9485.How do we know if these two numbers are included in those 3584 combinations?Please clear my doubt Mitch,these problems are giving me a hard time.



GMATGuruNY wrote:
dddanny2006 wrote:Can you please help me out with the Combinatorics method here.I want to know where I went wrong,as it can help me
Please see my corrections/notes in red:
dddanny2006 wrote:
Here's my method to do it.

Digits allowed-0,2,3,5,6,7,8,9

No of possible arrangements=8*8*8*8=4096

There are 4 conditions possible

1. All numbers are even =4*4*4*4=256

2. 3 numbers are even,one is odd =4*4*4*4=256
There are FOUR ways Case 2 could happen:
The 1st digit could be odd.
The 2nd digit could be odd.
The 3rd digit could be odd.
The 4th digit could be odd.
Thus, the result above must be multiplied by 4:
4*256.

3. 2 are even,2 are odd=4*4*4*4=256
There are SIX ways Case 3 could happen.
The following pairs of digits could be even:
1st and 2nd
1st and 3rd
1st and 4th
2nd and 3rd
2nd and 4th
3rd and 4th.
Thus, the result above must be multiplied by 6:
6*256.

4.1 is even,3 are odd=4*4*4*4=256
There are FOUR ways Case 4 could happen:
The 1st digit could be even.
The 2nd digit could be even.
The 3rd digit could be even.
The 4th digit could be even.
Thus, the result above must be multiplied by 4:
4*256.

Total good cases = 256 + (4*256) + (6*256) + (4*256) = 15*256.

P(good outcome) = (15*256)/4096 = 15/16.
Top
User avatar
GMAT Instructor
Posts: 1462
Joined: Thu Apr 09, 2015 9:34 am
Location: New York, NY
Thanked: 39 times
Followed by:22 members

by Jeff@TargetTestPrep » Mon Dec 18, 2017 11:30 am
dddanny2006 wrote:A four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?

a) ¼
b) ½
c) ¾
d) 15/16
e) 1/16
We can use the following equation:

P(at least one even digit) = 1 - P(no even digits)

Since 1 and 4 cannot be used, we have 8 available digits (0, 2, 3, 5, 6, 7, 8, 9), and we see that 4 of those 8 digits are odd (3,5,7,9). Thus, P(no even digits) = 4/8 x 4/8 x 4/8 x 4/8 = (1/2)^4 = 1/16.

Thus, P(at least one even digit) = 1 - 1/16 = 15/16.

Answer: D

Jeffrey Miller
Head of GMAT Instruction
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
Top
Post new topic Post Reply
• Page 1 of 1