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How many 4 digit numbers

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How many 4 digit numbers [#permalink]

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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
[Reveal] Spoiler: OA

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Re: How many 4 digit numbers [#permalink]

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abhisheksharma85 wrote:
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

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:
http://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 :-)
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Mike McGarry
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Re: How many 4 digit numbers [#permalink]

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New post 12 Nov 2015, 08:49
abhisheksharma85 wrote:
How many 4 digit numbers can be formed with the digits 0, 1, 2, 3, 4, 5, 6 and 6? (Repetition not allowed)

a. 220
b. 249
c. 432
d. 216
e. 288


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
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Re: How many 4 digit numbers [#permalink]

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New post 12 Nov 2015, 09:03
abhisheksharma85 wrote:
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


Taking Question with it's same language as given

To make number divisible by 5, the unit digit should be either 0 or 5 only

Case 1: If unit digit is fixed as zero

With two "6"s The choice to fill the remaining three digits = 3C2 x 5 = 15
With all remaining 3 digits different, The choice to fill the remaining three digits = 6 x 5 x 4= 120

Total Such cases = 120+15 = 135

Case 2: If unit digit is fixed as Five

With two "6"s and one "0" The choice to fill the remaining three digits = 2 [6605 or 6065]
With two "6"s and without "0" The choice to fill the remaining three digits = 3C2 x 4 = 12

With all remaining 3 digits different, and one "0" The choice to fill the remaining three digits = 2 (ways to place zero) x 5 x 4= 40

With all remaining 3 digits different, and without "0" The choice to fill the remaining three digits = 3 x 5 x 4= 60

Total Such cases = 2+12+40+60 = 114


Total numbers = 135+114 = 249

ANSWER OPTION B
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Re: How many 4 digit numbers   [#permalink] 12 Nov 2015, 09:03
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