Permutation and Combination.

[man111]

[tex](1)[/tex] how many natural numbers smaller than [tex]10^4[/tex] and divisible by [tex]4[/tex] can be formed using [tex]01235[/tex]

[tex](2)[/tex] find the sum of all numbers greater than [tex]10000[/tex] form by using [tex]02468[/tex] no digit being repeated.

[tex](3)[/tex] find the number of positive integers which can be formed by any number of digits [tex]012345[/tex] but using only once. how many

of these integers will be greater than [tex]3000[/tex]

[martin123456]

(1) i assume that u mean that we can use each digit not more than once
a number is divisible by 4 [tex]\Leftrightarrow[/tex] the number formed by the last to digits of its decimal representation is divisible by 4. From the given digits we can form the following two digits numbers that are divisible by 4: 12, 20, 32, 52. Now we have to count the number of the 3 digits numbers that are infront. We can generalize the counting process if we notice that 0 cannot be infront of a number that we are counting - the counting process can be generalized for numbers ending with 12, 32, 52 - let name these endings with xx. Also notice that no 1 can be infront also, for the counted number would be greater than 10000. With this in mind, we split the counting process again into 2 parts - the first is for 32, 52 in the end and the other concerns only 12 ending. In the first case we count _ _ _ xx. The first digit can be only one. The second can be choosen from 2 possible digits, the third one from one. So have 1.2.1=2 possibilities. The total count then is 2.2=4. Now we return back to the case 12. We count _ _ _ 12. The first digit can be choosen from 2 digits, the second - from 2 digits, and the third - from 1 digits. So the count is 2.2.1.=4. Up till now we have 8 numbers.
Finally, we return back to the case of 20. We count _ _ _ 20. For the first digit we can choose from 3 digits. For the second - from 3 digits, and for the last one - from 2. So the count is 3.3.2=18.
So totally we have counted 26 numbers.

[martin123456]

(2)
Since a natural cannot start with 0 and since we have exactly 5 digits we are assuming that every five digit number that we will sum will be greater than 10000.
The problem with the 0: replace the 0 with a digit x. We will sum the 5 digit numbers that can be formed using x2468 and from the calculated sum we will extract the numbers of the form 0 _ _ _ _. Think of the string [tex]\overline{abcde}[/tex] as of [tex]10^4a+10^3b+10^2c+10d+e[/tex].
The first sum: every digit can be infront and the last 4 digits give 4! permutations. So summing we will have [tex]10^4.4!(x+2+4+6+8)[/tex]. For the second summand - we get [tex]10^3.4!(x+2+4+6+8)[/tex] and so on. Totally we calculated [tex]10^4.4!(x+2+4+6+8)[/tex].
The second sum: 0 _ _ _ _. We have to extract the sum of the numbers _ _ _ _, formed by the digits 2,4,6,8. Deducting from above this sum is [tex]10^3.3!(2+4+6+8)[/tex].
Finally: [tex](2+4+6+8)(10^4.4!-10^3.3!)=20.10^3.3!(10.4-1)[/tex]

[martin123456]

(3)
one digit numbers are 5
two digit numbers can have first digit choosen from 5 digits, the 2nd - from 5, ...the count is 5.5
three digit numbers can be have first digit choosen from 5 digits, the 2nd - from 5 and the third from 4, so 5.5.4
four digit numbers - 5.5.4.3
five digit numbers - 5.5.4.3.2
six digit numbers - 5.5.4.3.2.1
so the count is A=5+5.5+5.5.4.3+5.5.4.3.2+5.5.4.3.2.1
we extract the count of 1,2, 3digit numbers - so we totally extract B=5+5.5+5.5.4.
the number we look for is A-B - the number of number that are in the interval 1000 to 3000 that are formed from the above digits. Notice that we can't form a the number 3000.
the first digit is 1 or 2. Then we have 5 possibilities for the second digit, 5 for the 3rd, 4 for the fouth. Totally 5.5.4 multiplied by 2, so 5.5.4.2.