medium hard permutations

[Night reader]

How many four-digit odd numbers do not use any digit more than once?

(A) 1728
(B) 2160
(C) 2240
(D) 2268
(E) 2520

[shovan85]

-- Deleted --

[Rahul@gurome]

How many four-digit odd numbers do not use any digit more than once?
(A) 1728
(B) 2160
(C) 2240
(D) 2268
(E) 2520

Let the number be ABCD.

D can be filled in 5 ways (1,3,5,7,9), as the number is odd.
A can be filled in 8 ways (1-9 but not the ones used in D; First digit can't be zero)
B can be filled in 8 ways (0-9 but not the ones used in C and D)
C can be filled in 7 ways (0-9 but not the ones used in B, C and D)

Thus total number of such numbers = 5*8*8*7 = 2240

The correct answer is C.

[goyalsau]

I thought there will be same number of odd and even 4 digit Numbers.

first Digit Can be filled in 9 ways,
Second in 9 Ways,
Third in 8 ways
Forth in 7 ways,

9 * 9 * 8 * 7 = 648 * 7 = 4536

4536 / 2 = 2268 Answer is 2240,

I thought Its good if i count Number of Even Digits The Method by which Rahul Has Counted Odd Digits
But again answer is 2240, Which is again Wrong. Because No. of Even Digits Must be 4536 - 2240 = 2296


Now I have counted 56 Less digits than required. Why is that so, I was almost sure that it is because of our own Arya Bhatt Zero

Then i counted Them With or Without Zero.
ABCD.

I ) with Zero. D = 1 ( Zero of D )
A = 9 , B = 8 , C = 7 .
Number of digits with Zero Must be 9 * 8 * 7 = 504.

II ) Without Zero. D = 4 ( ONLY 2, 4, 6, 8 )

A = 8 , B = 8 ( Zero Can Come at B ) , C = 7

8 * 8 * 7 * 4 = 1792

So total number of Even Digits must be 504 + 1792 = 2296 ,

Once WE know Number of Even Digits Then its Almost Done Total - Even = ODD4536 - 2296 = 2240

Which is Mistakenly Correct.

[swathi8388]

Though i understood the answer I am confused why I wont get the Ssame answer the other way..that is..find the total number 4 digit numbers that have unique digits and divide it by 2 (cause in any set of numbers the odd and even numbers have to be equal?)

but in that case we get 2268 :

9 * 9 * 8 * 7 = 648 * 7 = 4536

4536 / 2 = 2268

can someone explain.

[Brent@GMATPrepNow]

Though i understood the answer I am confused why I wont get the Ssame answer the other way..that is..find the total number 4 digit numbers that have unique digits and divide it by 2 (cause in any set of numbers the odd and even numbers have to be equal?)

but in that case we get 2268 :

9 * 9 * 8 * 7 = 648 * 7 = 4536

4536 / 2 = 2268

can someone explain.

You are assuming that the 4536 numbers are such that half are odd and half are even. This is not the case.
Since we are, in essence, saying that the thousands digit cannot be zero (an even number), then we are increasing the likelihood that zero appears among the other 3 digits (including the units digit).
So, among those 4536 numbers, more than half are even, which means fewer than half are odd.
So, when you divided 4536 by 2, the result is too big.

Cheers,
Brent

[Brent@GMATPrepNow]

How many four-digit odd numbers do not use any digit more than once?

(A) 1728
(B) 2160
(C) 2240
(D) 2268
(E) 2520

Rahul's solution above is great. It applies something called the By the Fundamental Counting Principle (FCP). This technique is also called the Slot Method.

It's important to note, that when you apply this strategy, you should always begin with the most restrictive cases

Here's what I mean.

We're going to take the task of building acceptable 4-digit numbers and break it into stages.
Our stages will consist of selecting the four digits needed. Among these stages, the most restrictive one is the units digit, since it must be odd. So, we'll begin there.

Stage 1: Select the units digit
Since the digit must be 1, 3, 5, 7, or 9, we can complete this stage in 5 ways.

From here, the 2nd most restrictive stage is selecting the thousands digit (since it cannot be zero). So . ..

Stage 2: Select the thousands digit
This digit cannot equal zero and it cannot be the same as the digit selected as the units digit.
This leaves 8 digits to choose from, so we can complete this stage in 8 ways.

At this point, the two remaining stages are equally restrictive, so we can complete either stage next.

Stage 3: Select the tens digit
This digit cannot equal either of the 2 digits already-selected digits.
This leaves 8 digits to choose from, so we can complete this stage in 8 ways.

Stage 4: Select the hundreds digit
This digit cannot equal either of the 3 digits already-selected digits.
This leaves 7 digits to choose from, so we can complete this stage in 7 ways.

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus build a 4-digit integer) in (5)(8)(8)(7) ways ([spoiler]= 2240 ways[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775