6-digit Arrangements

First we have to know that numbers divisible by $4$ have to have two last algarisms divisible by $4$ . Then the possibilities to this is ${12,16,24,32,36,52,56,64 = 8}$ possibilities. However are still missing the four firsts algarisms that be organized by $4!$ differents ways, then we have $8 * 4! = 192$ . But our amostral space is $6!$ that are the all forms to creat differents six algarisms numbers using $1,2,3,4,5,6$ . In this way, we have the probability is $P = \frac{192}{720} = \frac{4}{15}$ . The answer is $4+15 = 19$ .

1. Divisibilty by 4: Number must be finished (in this case): $(12, 16, 24, 32, 36, 52, 56, 64) \rightarrow 8$ cases..
2. When we fill 2 last numbers, remain 4 numbers to permutation.
3. $P_{4}$ : $4!$
4. All numbers wanted: $8 \cdot 4!$

If the last two digits are divisible by 4, the whole number will be divisible by 4. The only possible endings for the six digit numbers will thus be the numbers 12, 16, 24, 32, 36, 52, 56 and 64.

With these eight endings, there will only by four numbers left to place, giving in total

$(4 \times 3 \times 2) \times 8 = 192$

numbers divisable by 4.

There are $6! =720$ possible numbers in all, giving the probability as

$\frac{192}{720}=\frac{4}{15}$

$4+15= \boxed{19}$

The total no. of 6-digit nos. using these 6 digits=6!=720;

For a no. to be divisible by 4,last 2 digits should be divisible by 4.Thus the last two digits can be 12,16,24,32,36,52,56 or 64.The no. of required nos. with two digits fixed=4!=24;

Thus,total no. of nos. divisible by 4=8*24=192;

Thus probability=192/720=4/15;

4+15=19.

The number would be divisible by $4$ if and only if last $2$ digits of it are divisible by $4$ . Now consider possible two-digit numbers which can be formed using digits from the set $(1,2,3,4,5,6)$ using one element at a time. These are: $(12,16,24,32,36,52,56,64)$ . Thus there are $8$ such digits. Now if we choose any of these as last two digits, remaining four digits can be permuted in $4!$ ways. Also, without any restriction, we can form $6!$ total numbers. Hence the required probability is:

$\frac{4!*8}{6!}=\frac{4}{15}$ .

Hence the required answer is: $4+15=\boxed{19}$

In order to check the divisibility of $4$ of a number, we check the last $2$ digits. If it is divisible by $4$ , then the whole number is divisible by $4$ , and also converse. So, let's have a look on how many 2-digits number can be divided by $4$ to make our life easier later on. The numbers which are multiples of $4$ are $12, 16, 24, 32, 36, 52, 56, 64$ . We ignore $44$ since it is mentioned that each digit is only used once in forming the 6-digit number. From here, we know that there are only 4 spaces if we put any number listed above as the last $2$ digits. Therefore, for each case, we have $4! = 24$ possibilities, because of that we have $192$ numbers that is divisible by 4. Then the amount of the numbers that can be formed by using $1, 2, 3, 4, 5, 6$ is $6!=720$ . Thus, the possibility would be $\frac{192}{720}$ $=$ $\frac{4}{15}$ . Comparing with the fraction $\frac{a}{b}$ , we get $a=4$ and $b=15$ . So, $a+b = \boxed {19}$

Numbers that are divisible by 4 are any number has the last two digits divisible by 4. for these six digits numbers are 12,16,24,32,36,52,56,64 (8 numbers)

so we have 8*(4P4)=192 numbers are divisible by 4 and we have 6P6=720 numbers could be formed from 6 digits

so the probability is 192/720=4/15 so a+b= 19

The digits 1, 2, 3, 4, 5, 6 can be arranged in 6! = 720 ways. Now, when the last 2 digits of the number is divisible by 4 then the whole number is divisible by 4. The possible last 2 digits, divisible by 4 are 12, 16, 24, 32, 36, 52, 56, 64. Now, the remaining 4 digits can be arranged in 4! = 24 ways. The required probability is (8 × 4!) / 720 = 192 /720 = 4/15. The required answer is (4+15) = 19