We have six digit-number* and we need to place in it digits 1, 2 and 3 in free order. We can do it in 6*5*4 ways, the remaining 3 places may include any digit, so it gives us 6*5*4*10*10*10 different numbers.
I'm really stuck on this problem.
It says,how many numbers between 1 and 999,999 contains each of the digit 1,2 and 3 at least once.(Hint : for each i = 1,2,3 A_i be the set of all integers from 1 to 999,999 that do not contain the digit i)
I really need help with this one.
Thanks for your help Plato.I only understand parts of it.Like,
A(1) = the number of integers that contain a 1 as one of it's digit.
A(2) = the number of integers that contain a as one of it's digit.
and so is A(3).
now A(1) = _ _ _ _ _ _
and each of these blanks cna be filled with a any of the digits between 0 to 9
i.e we have 10 choices
so if it's 0 0 0 0 0 1
0 1 9 9 4 5
and so on
so,A(1) has to have a digit 1 in it.
_ _ _ _ _ _ = 1 * 10 * 10 * 10 * 10 * 10
isn't it? how do u say it's 10 ^6 -1 then?
I was using the hint given in the problem.
is the number of numbers from 1 to 999999 that have no 1’s.
That is : you can use any of the other nine digits, subtract 1 to account for 000000.
is the number of numbers from 1 to 999999 that have no 1’s nor 2’s.
That is so forth.
The total number of numbers from 1 to 999999 is .
Now use inclusion/exclusion to get what you want.
Try shorter way: number xxxxxx has to include each of 1, 2 and 3 digits, they can be placed in different ways:
xxx321
xx3x21
x3xx21
3xxx21
xxx231
xx32x1
x3x2x1
3xx2x1
xx2x31
xx23x1
...
1x32xx
13x2xx
1x2xx3
1x2x3x
1x23xx
132xxx
12xxx3
12xx3x
12x3xx
123xxx
Each x may be replaced by any other digit, so each of 120 different cases represent 10 different numbers, that's why we get