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

2 Answers

Answer:

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#

Oct 8, 2017

Answer:

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).