A ten digit number is formed using the digits from zero to nine,every digit being used only once.find the probability that the number is divisible by 4

For the number to be divisible by 4, the last two digits should be divisible by 4.

These include digits ending with:
00
04
08
12
16
20
24
28
32
36
40
44
48
52
56
60
64
68
72
76
80
84
88
92
96

Now, since the digits are used only once, there are 'less' numbers to pick from:
04
08
12
16
20
24
28
32
36
40
48
52
56
60
64
68
72
76
80
84
92
96

The total number of digits is given by 10! - 9! = 3265920

Of these digits, those that end in the numbers I mentioned above are divisible by 4.

Can you continue now? :)

The total number of digits is given by 10! - 9! = 3265920

Of these digits, those that end in the numbers I mentioned above are divisible by 4.

Can you continue now? :)

Unk - maybe I don't quite follow you, but it seems to me that since you showed that there are 22 ways that the last two digits make a mumber divisible by 4, the probability that a number is divisible by 4 is simply 22 divided by the number of possibilities for the last two digits, which is 10 x 9.

Hm...

10 digits, total number of digits = 10!
Removing numbers starting with zero, get get = 10! - 9!

Now, for numbers ending with 04, there are 8! Numbers.
That goes on 5 times, for numbers with 0 in the last two digits.
So, 5 x 8! Numbers up to now.

Then, 8! - 7! Numbers ending with 12.
This goes on for 17 numbers, hence, 17(8! - 7!)

For a total of 17(8! - 7!) + 5(8!)

Then, let X be the number.

P(X\ is\ divisible\ by\ 4) = \frac{17(8! - 7!) + 5(8!)}{10!-9!} = \frac{53}{216}

or am I missing something? :(

Ahh.. my way assumed that any digit can be in any position, whereas your technique assumes that 0 is not allowed to be the first digit. This slightly raises the probability of 0 appearing in one of the last two places, which raises the probability of the number being divisible by 4. So whereas I get 22/90 = 0.2444.. you get 53/216 = 0.24537...

Sirinivas - the answer depends on whether a number is allowed to begin with the digit 0.

For the number to be divisible by 4, the last two digits should be divisible by 4.

These include digits ending with:
00
04
08
12
16
20
24
28
32
36
40
44
48
52
56
60
64
68
72
76
80
84
88
92
96

Now, since the digits are used only once, there are 'less' numbers to pick from:
04
08
12
16
20
24
28
32
36
40
48
52
56
60
64
68
72
76
80
84
92
96

The total number of digits is given by 10!-9!=3265920 [Excluding those starting with a 0]

Of these digits, those that end in the numbers I mentioned above are divisible by 4.

Among the above mentioned 22 cases there are 16 cases where the arrangement is starting with a '0'.
So, 16*7! Cases should be excluded from 22*8! Cases.

Hence, the probability that the number is divisible by 4 is
P = (22*8! - 16*7!) / (10! - 9!) = 20/81 [Ans.]