[sirinivas]

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

[Unknown008]

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

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

Can you continue now? :)

[ebaines]

The total number of digits is given by

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.

[Unknown008]

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.



or am I missing something? :(

[ebaines]

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.

[thegreymatter]

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.]