Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
09 Oct 2013, 10:20
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
43% (02:19) correct
57%
(02:04)
wrong
based on 28
sessions
History
Date
Time
Result
Not Attempted Yet
How many 4 digit numbers can be formed with the digits 0, 1, 2, 3, 4, 5, 6 and 6?
a. 220
b. 249
c. 432
d. 216
e. 288
a. 220
b. 249
c. 432
d. 216
e. 288
This Question is Locked Due to Poor Quality
Hi there,
The question you've reached has been archived due to not meeting our community quality standards. No more replies are possible here.
Looking for better-quality questions? Check out the 'Similar Questions' block below
for a list of similar but high-quality questions.
Want to join other relevant Problem Solving discussions? Visit our Problem Solving (PS) Forum
for the most recent and top-quality discussions.
Thank you for understanding, and happy exploring!
09 Oct 2013, 14:30
Kudos
Bookmarks
abhisheksharma85
Dear abhisheksharma85,
I'm happy to comment on this.
What is the source of this question?? From what I can tell, not only is the OA not correct, but the answer I calculate isn't listed among the answer choices and isn't even close. This is a fascinating counting question. For more on counting, including the Fundamental Counting Principle, see:
https://magoosh.com/gmat/2012/gmat-quant-how-to-count/
Here's how I would approach it. First, I'll ignore the repeat, and count the four-digit numbers I can make with {0, 1, 2, 3, 4, 5, 6}. I am assuming that, in order to be a true four-digit number, zero cannot be in the first digit, the thousands place --- i.e. 0246 does not count as a "four-digit" number.
For the four-digit numbers from {0, 1, 2, 3, 4, 5, 6} ---
For the thousands digit, {1, 2, 3, 4, 5, 6} --- six choices
For the hundreds digit, drop the one picked in the first choice, but add 0 as a possibility --- still six choices
For the tens digit, now also drop what was picked in the hundred digit --- now, five choices
For the ones digit, now also drop what was picked in the tens digit --- now, four choices
Total number = 6*6*5*4 = 720
For example, first choice, from six options {1, 2, 3, 4, 5, 6}, I could choose 3
Then, from these six options {0, 1, 2, 4, 5, 6}, I could choose 0
Then, from these five options {1, 2, 4, 5, 6}, I could choose 2
Then, from these four options {1, 4, 5, 6}, I could choose 1
This produces the unique four-digit number 3021
Now, we have to consider the four-digit numbers with two 6's. First, let's think of where the two 6's could fall among the two digit --- there are 4C2 = 6 possible locations for the two 6's:
(a) 66 _ _
(b) 6 _ 6 _
(c) 6 _ _ 6
(d) _ 66 _
(e) _ 6 _ 6
(f) _ _ 66
I grouped them this way, because in (a)-(c), the thousands digit is already occupied, so zero would be one legitimate choice for either of the other slots, BUT in (d) - (f), we have to be careful, again, not to place zero in the blank in the thousands place.
For each of (a) - (c), we have six choices {0, 1, 2, 3, 4, 5} for the left slot, and then, dropping that one digit, five choices remaining for the right slot. 5*6 = 30 for each of the three, so this results in 90 more numbers.
For each of (d) - (f) we have five choices {1, 2, 3, 4, 5} for the left slot, the thousands place; then, for the right slot, we drop the digit we chose already, but we add zero as a possible choice, and thus we still have five choices. 5*5 = 25 for each of the three, so this results in 75 more numbers.
Altogether 720 + 90 + 75 = 885
That's my count of the number of possible four-digit numbers we could form from this set.
Does all this make sense?
Mike
General Discussion
12 Nov 2015, 08:49
Kudos
Bookmarks
abhisheksharma85
Assuming that the repetition of digits is not allowed
Case 1: If unit digit is fixed as zero
Then the choices at the remaining three places are 6 x 5 x 4 = 120
Case 2: If unit digit is fixed as Five
Then the choices at the remaining three places are 5 x 5 x 4 = 100
Total Such number = 120+100 = 220
ANSWER: OPTION A
