using these digits 0, 1, 2, 3, 4, ,5 , 6, 7. How many six digits even numbers can be formed???

Question

i forgot to add more details. Number can not be repeated

Sorry!!

Answer 1

As the no. is even so 6th digit has to be either 0 or 2 or 4 or 6.

When 6th digit is 0,
5th can be anyone from 1 to 7.
4th can be anyone from 1 to 7 except the 5th digit.
3rd can be anyone from 1 to 7 except the 5th and 4th digit.
........
1st can be anyone from 1 to 7 except 5th, 4th, 3rd and 2nd digit.


So there can be 7*6*5*4*3 = 2520 no. of 6-digit even numbers when last digit is 0.

When 6th digit is 2.
5th can be anyone from 0 to 7 except 2.
4th can be anyone from 0 to 7 except the 5th digit and 2.
3rd can be anyone from 0 to 7 except 2, 5th and 4th digit.
........
1st can be anyone from 0 to 7 except 2, 5th, 4th, 3rd and 2nd digit AND 0.

So there can be 7*6*5*4*2 = 1680 no. of 6-digit even numbers when last digit is 2.

Similarly, there can be 1680 no. of 6-digit even numbers when last digit is 4 or 6.

So total no. of 6-digit even number possible with the given digits is (2520 + 3*1680) = 7560.
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@ Optimizer

A 6-digit number cannot have the 1st digit equal to 0 (zero) as then it becomes a 5-digit number.

Answer 2

This is moderately complicated, from the fact that you cannot use the digit 0 as the first digit, assuming that by a 6 digit number one means a proper 6 digit number, not something like 025738... So to correctly answer this question, we need to break it down into two possibilities:

1) The last digit is 0. In this case we need only select a string of 5 digits from 1,2,3,4,5,6,7 to be the first 5 digits of the number. There are P(7,5) ways of doing this (I am assuming here that you know what that means)
P(7,5)=7!/2!=7*6*5*4*3=2520

2) The last digit is not 0, so there are three equally likely possibilities for the last digit, either a 2,4 or 6. For each of these possibilities, we need to choose a non-0 first digit, and there will be 6 possibilities for that (for example, if the last digit were a 4, then the possibilities for the first digit would be 1,2,3,5,6,7) and for the middle 4 digits we would need to choose a string of 4 digits from the remaining 6 digits available. Thus the number of possibilities for this case would be 3*6*P(6,4) where the three is counting the possibilities for the final digit, the 6 is counting the possibilities for the first digit and the P(6,4) is counting the possibilites for the middle 4 digits. So the total for this case would be 3*6*6!/2!=3*6*6*5*4*3=6480.

Finally to get the total number of possibilities, we would add the numbers from the separate cases: 2520+6480=9000

Answer 3

The answer is 9000.

There are 4 possible digits to be the 6th place: 0, 2, 4, 6.
Whichever of these we pick, there are 7 possible digits left for the 5th place.
Then, there are 6 possible digits left for the 4th place,
5 left for the 3rd place,
4 left for the 2nd place,
and 3 left for the first place.

This gives 4 x 7 x 6 x 5 x 4 x 3 = 10,800 possible numbers.

But I have counted numbers starting with 0, as well, which shouldn't be allowed. I must subtract those off. How many even numbers starting with zero can be formed with these digits?

We fix zero as the first digit.
Then there are 3 possible digits for the 6th place 2, 4, and 6.
Whichever of these we pick, there are 6 possible digits left for the 5th place (not 0, and not whatever was in the 6th place).
Then there are 5 digits left for the 4th place,
4 digits left for the 3rd place,
3 digits left for the 2nd place.

This means there are 3 x 6 x 5 x 4 x 3 = 1,080 even 6-digit "numbers" which start with zero.

So the total number of possible six-digit even numbers is 10,080 - 1,080 = 9,000.

I wrote a program to generate all such numbers to verify that my answer was correct. You can see all 9000 of them at http://www.math.purdue.edu/~dbabcock/ya/sixdigitevens.txt .

Answer 4

In base 8, the smallest six digit number is 100,000, the largest is 777,777. Thats from 8^5 to 8^6 - 1. Half of that are even numbers, so the answer, including 100,000, is (1/2)(8^6 - 8^5), or 114,688.

For comparsion, if the digits included 8 and 9, then we have numbers from 100,000 to 999,999, or 899,999. Including 100,000, we have (1/2)900,000 = 450,000.

Oh, the digits cannot be repeated? Okay, if the last digit is either 2,4,6, then the total number is 3*6*6*5*4*3 = 6480, and if the last digit is 0, then the total number is 1*7*6*5*4*3 = 2520, so that we have 6480 + 2520 = 9000 exactly. So, I guess the other guys beat me to this.

Answer 5

There are 8 digits, of which four (0,2,4,6) are legal 'right-most' digits for an even number. For each of the four cases where one of these 'even' digits is pinned in the rightmost position, there are 7P5 = 7!/2! = 7*6*5*4*3 permutations of the remaining seven digits to chose for the remaining 5 slots of the number.

So the answer is (4)(7! / 2!) = (4)(2520) = 10080 such even numbers
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(EDITED: pshowmick has it correct below; I forgot to allow for the fact that the digit 0 cannot be in the left-most position for a true 6-digit number)

Answer 6

4x7x6x5x4x3 -3x6x5x4x3x2= 7920