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Number of 7-digit codes from 1,234,567

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Number of 7-digit codes from 1,234,567

by gmattesttaker2 » Sat Apr 19, 2014 10:13 pm
Hello,

Can you please tell me if my approach is correct here:

How many 7-digit codes can be made by rearranging all the digits of the number
1,234,567, if the thousands digit must be odd?

(A) 1,080
(B) 1,440
(C) 2,160
(D) 2,880
(E) 3,120

Thousands digit could be 1 or 3 or 5 or 7. Hence number of 7 digit codes is:

7 x 7 x 7 x 4 x 7 x 7 x 7

However, OA is given as D


Thanks a lot,
Sri
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by [email protected] » Sat Apr 19, 2014 10:47 pm
Hi Sri,

I'm going to give you a hint and let you try this question again. According to the instructions in this question, you must use each digit ONCE. Think about how that would affect the "math."

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by gmattesttaker2 » Sat Apr 19, 2014 11:12 pm
[email protected] wrote:Hi Sri,

I'm going to give you a hint and let you try this question again. According to the instructions in this question, you must use each digit ONCE. Think about how that would affect the "math."

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Rich
Hi Rich,

I was just wondering why we cannot repeat the digits since the question doesn't mention anything?

Thanks a lot,
Sri
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by [email protected] » Sun Apr 20, 2014 12:14 am
Hi Sri,

The prompt tells us that we have to "rearrange" the digits, implying that each digit is used just once. Attention to these types of details matters....

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by theCodeToGMAT » Sun Apr 20, 2014 12:28 am
1, 2 3 4, 5 6 7

_, _ _ X, _ _ _

ODD = 1, 3, 5, 7

Considering one ODD Number, say 1 = ways = 6!

Since, this is valid for all 4 ODD ..

So, 4 * 6! = 4 * 720 = 2880

[spoiler]{D}[/spoiler]
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by GMATGuruNY » Sun Apr 20, 2014 2:14 am
gmattesttaker2 wrote:How many 7-digit codes can be made by rearranging all the digits of the number
1,234,567, if the thousands digit must be odd?

(A) 1,080
(B) 1,440
(C) 2,160
(D) 2,880
(E) 3,120
Start with the MOST RESTRICTED POSITION.

Here, the most restricted position is the thousands digit, which must be ODD.
Number of options for the ODD thousands digit = 4. (1, 3, 5, or 7.)
Number of ways to arrange the remaining 6 digits = 6!.
To combine these options, we multiply:
4 * 6! = 2880.

The correct answer is D.
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by Brent@GMATPrepNow » Sun Apr 20, 2014 8:40 am
gmattesttaker2 wrote: How many 7-digit codes can be made by rearranging all the digits of the number
1,234,567, if the thousands digit must be odd?

(A) 1,080
(B) 1,440
(C) 2,160
(D) 2,880
(E) 3,120
My approach is the same as some others in that it employs the Fundamental Counting Principle (FCP).
I just to formalize the approach so that people can see why the answer is D

Take the task of arranging the 7 digits and break it into stages.

We'll begin with the most restrictive stage.

Stage 1: Select the thousands digit
Since this digit must be ODD, we can choose from 1, 3, 5 or 7
So, we can complete stage 1 in 4 ways

Stage 2: Select the units digit
There are 6 remaining digits from which to choose, so we can complete this stage in 6 ways.

Stage 3: Select the tens digit
There are 5 remaining digits from which to choose, so we can complete this stage in 5 ways.

Stage 4: Select the hundreds digit
We can complete this stage in 4 ways.

Stage 5: Select the ten thousands digit
We can complete this stage in 3 ways.

Stage 6: Select the hundred thousands digit
We can complete this stage in 2 ways.

Stage 7: Select the millions digit
We can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete all 7 stages (and thus arrange all 7 digits) in (4)(6)(5)(4)(3)(2)(1) ways ([spoiler]= 2880 ways[/spoiler])

Answer: D
--------------------------

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