Question

How many 6 digit combinations can i get using numbers 1-49?

Answer 1

No. of possible combinations without repeating any number `=43,084`

Explanation:

Four combinations 6 digits are
1) All six digits are single digits 1 to 9.
`9C6=(9!)/((6!)(9-6)!)=(9*8*7)/(1.2.3)=84`

2) All 6 digits are 3 double digits 10 to 49.
`40C3=(40!)/((37!)(3!))=(40*39*38)/(1*2*3)=9880`

3) Two single digits and two double digits.
`=((9!)/((2!)(7!)))*((40!)/((2!)(38!)))`
`=((9*8)/(2))*((40*39)/(2))=36*780=28080`

4) Four single digits and one double digit.
`=((9!)/((4!)(5!)))* ((40!)/((1*)(39!)))`
`=((9*8*7*6)/(1*2*3*4))*(40)=126*40=5040`

Adding all 4 options, we get the answer
`=84+9880+28080+5040=43,084`

Answer 2

Answer is `43,084`.

Explanation:

There are four overall possibilities.

You can get a six digit number only if you take `rarr`

(1) 6 single digit numbers.
(2) 3 double digit numbers.
(3) 4 single & 1 double digit numbers.
(4) 2 single & 2 double digit numbers

Now, from `(1,2,.....,48,49)` `rarr`
There are `9` single & `40` double digit numbers.

So, taking care of all these cases as well as the probabilities of the position of all the numbers in the six digit number we get`rarr`

`P=9C_6+40C_3+9C_4xx40C_1+9C_2xx40C_2`

`:.P=84+9,880+5,040+28,080`

`:.P=43,084`.

Therefore, total number of required combinations is `43,084`. (Answer).