Question

Combinations
I must find how many 7 digit registration numbers can start with 1, when the digits available are 0, 1, 2, 3, 4, 5, 6, 7, 9, the registration numbers are issued in ascending order and you cannot have 6 and 9 in the same line,

Answer 1

looks same q ?

Answer 2

you fixed first space wid 1, you're left wid 6 spaces to fill - and you have a constraint that 6 and 9 cannot coexist. Is that right ?

Answer 3

First, calculate the number of combinations without including the 6th digit.

Answer 4

yes that is right

Answer 5

yes thats a good idea, first calculate number of permutations without 6 digit

Answer 6

k

Answer 7

c= 7!/(7-6)! c = 5040

Answer 8

Correct ! thats the total permuations with out digit 6.

Answer 9

k

Answer 10

Next, find total permutations with out 9, and with 6
those will be c = 7!/(7-6)! = 5040

but there are some repititions here

Answer 11

how do u get rid of the repetitions.

Answer 12

repititions = permutations that doesnt have both 6 and 9 = 6!

Answer 13

so it is the same?

Answer 14

so total permutations starting wid 1 w/o 6 and 9 coexisting = 5040 + 5040 - 6!

Answer 15

this is same as the solution we worked in your previous q, 7! + 6x6!

Answer 16

here:

Answer 17

dont ignore the factorial sign \(\color{red}{!}\)

Answer 18

oh yeah oops

Answer 19

ha otherwise, it looks good.. replace 6! with 720

Answer 20

c=9360

Answer 21

9360 = \(\large \color{red}{\checkmark}\)