If 6 different numbers are to be selected from integers 0 to 6, how ma

Question

If 6 different numbers are to be selected from integers 0 to 6, how many 6-digit even integers greater than 300,000 can be composed?

OA:  1680

fluke · Answer

Use the slot method:

(3)()()()()()

Let's fix 3 @ Hundred thousand place, which gives us just 1 option
Units place can be: 0,2,4,6(Total=4 options)=>(1)()()()()(4)
Now, 2-digits out of 7(0,1,2,3,4,5,6) are already taken; Thus, remaining can be 5;4;3;2
Tens=5: (1)()()()(5)(4)
Hundreds=4: (1)()()(4)(5)(4)
Thousands=3: (1)()(3)(4)(5)(4)
Ten Thousands=2: (1)(2)(3)(4)(5)(4)
Likewise for odd 5 @ Hundred thousand place.
Thus, for odds (3,5)@ Hundred thousand place: (2)(2)(3)(4)(5)(4)

However, for evens 4,6 @ Hundred thousand place, we will just have three options for units place(0,2,6) OR (0,2,4)
Options will look like: (1)(2)(3)(4)(5)(3)
Thus, for evens (4,6)@ Hundred thousand place: (2)(2)(3)(4)(5)(3)

Overall:
(2)(2)(3)(4)(5)(4)+(2)(2)(3)(4)(5)(3)=960+720=1680

Ans: "1680"

Raghava747 · Answer

Hello ,
There are 1680 numbers possible

What is OA

My explanation is as follows,

Greater than 300,000: the possible numbers are 300,002 to 600,000+

The following cases need to be considered

Starting with number n >= 3and last number = {0,2,4,6}

case I :
Even number
4 or 6 in 100 thousands digit

the units digit can be any one of {0,2,4(if 6 is considered), 6 (if 4 is considered)}

so IF 4 is considered in 100 thousands digit

1*5*4*3*2*3 = 360

So if 6 is considered in 100 thousands digit

1*5*4*3*2*3 = 360

Hence in Case I we have 720

Case II
3 or 5 in 100 thousands digit

So if 3 is considered in 100 thousands digit

1*5*4*3*2*4 = 480

So if 5 is considered in 100 thousands digit

1*5*4*3*2*4 = 480

Hence For case II we have 480+480= 960

Total we have 960+720 = 1680 cases:

Consider Kudos if my explanation was helpful. Its a great option on GMATClub: :-D

shahideh · Answer

Consider the number A B C D E F

If 5 is placed in A digit: 
we have 1 possibility for A digit
4 possibilities for F digit
and 5, 4, 3, 2 possibilities for B, C, D, and E.
So, 1*5*4*3*2*4=480

If 4 or 6 is placed in A digit :
we have 2 possibility for A digit
3 possibilities for F digit
and 5, 4, 3, 2 possibilities for B, C, D, and E.
So, 2*5*4*3*2*3=720

If 3 is placed in A digit: 
we have 1 possibility for A digit
4 possibilities for F digit
and 5, 4, 3, 2 possibilities for B, C, D, and E.
So, 1*5*4*3*2*4=480

Total=480+720+480=1680

EMPOWERgmatRichC · Answer

Hi All,

While this is an old question, there's a broader lesson in the way the post was presented. We're essentially 'stuck' doing math because the original poster didn't include the 5 answer choices. On Test Day, you will face many questions in which the 'format' of the answer choices can provide you a huge hint (or shortcut) as to how you can go about solving the problem.

In this particular prompt, we're asked to determine the total number of possible 6-digit numbers that fit a certain series of restrictions. One of those restrictions is that we must use only the digits 0 to 6 (inclusive) without any duplicates. IF that was the only restriction, then there would be....

(6)(6)(5)(4)(3)(2) = 4320 options (since the 6-digit number cannot start with a 0).

Since there are additional restrictions, the correct answer would clearly be less than 4320. IF the answer choices included just one answer that was less than 4320, then that answer would have to be the correct one and we wouldn't have to do any more work to get to the solution.

GMAT assassins aren't born, they're made,
Rich

bumpbot · Answer

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If 6 different numbers are to be selected from integers 0 to 6, how ma

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