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

Oct 7, 2017

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!)))
$= \left(\frac{9 \cdot 8}{2}\right) \cdot \left(\frac{40 \cdot 39}{2}\right) = 36 \cdot 780 = 28080$

4) Four single digits and one double digit.
=((9!)/((4!)(5!)))* ((40!)/((1*)(39!)))
$= \left(\frac{9 \cdot 8 \cdot 7 \cdot 6}{1 \cdot 2 \cdot 3 \cdot 4}\right) \cdot \left(40\right) = 126 \cdot 40 = 5040$

$= 84 + 9880 + 28080 + 5040 = 43 , 084$

Oct 8, 2017

Answer is $43 , 084$.

#### Explanation:

There are four overall possibilities.

You can get a six digit number only if you take $\rightarrow$

(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 $\left(1 , 2 , \ldots . . , 48 , 49\right)$ $\rightarrow$
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$\rightarrow$

$P = 9 {C}_{6} + 40 {C}_{3} + 9 {C}_{4} \times 40 {C}_{1} + 9 {C}_{2} \times 40 {C}_{2}$

$\therefore P = 84 + 9 , 880 + 5 , 040 + 28 , 080$

$\therefore P = 43 , 084$.

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