How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?
Any help will be appreciated.
a number of for digit is someting who is wrote
abcd where those for letter are any decimal digit (0-9) you want
(when a=0 or ab=00 this may not be considered as four digit number is up to you to check or to prcise in your answer the choice you make)
first lets form numbers whith four diferent digits
you chose a youve got 10 (or 9 if you consider the remark the contrary of below) then you choose b you have now 9 possibility then c 8 possibility then d 7 possybility left then youve got one number whith differents digits
i know because i have done this sort of exercise a lot of time (but i admit i have no formal explanation to give you)
that they are then 10*9*8*7 numbers of four different digits that you can form
try whith two digit whit less digits ( 0 and 1) ect and get your own explanation!!
then try to find the proportion of numbers between oo and 99 that satisfy the odd even relation you mentioned
as they are as many numbers between 3500 and 3600 finishing by eleven
than numbers between 4700 and 4800 finishin by seventeen you will have no problem (not real anyway ) to conclude (i know i should have put eleven instead of seventeen but it would be no joke!)
0 1 2 3 4 5 6 7 8 9
Note that 5 of these digits are even and five are odd.
we want a four digit number where the first digit is odd, the second digit is even and all digits distinct. Here's how we get such a number.
For the first digit, we have 5 choices, since there are 5 odd digits.
For the second digit, we also have 5 choices, since we have 5 even digits still
For the third digit, we have 8 choices, since we already choose 2 digits that we can't choose again
For the fourth digit, we have 7 choices since we chose 3 of the 10 digits and we can't choose them again
So all four digit numbers of this form is given by the product of all the choices. Hence your solution is:
5 * 5 * 8 * 7 = 1400 possible 4-digit numbers of this form
i realized my answer was wrong after posting it:
of course when you have chosen the first two numbers ab of abcd
you have not the same set of possybility for c and d so the proportion is not always the same
my litle finger told me that you have to consider 3 possibility for a and b : their all even their all odd and the other possibility i suppose the numbers of good combination for c and d depends only on that mater and nothing else
so you would had to count how many numbers ab belongs to wich case ect...
they may be also a probleme whith a=0 is valid or not (but as you have a choice)
they must be other solution as well (the contrary would be a real absurdity)
sory i rode firt in stead of 'leftmost' because lefmost is not a real word so thats how my (our) brain is(are) functioning
any way i am very sory to post again and not editing my post (it is not so bad because my post are generaly folowing one an other) but when you say a big mistake (and see it just after sending it ) it is not good !
so if somebody reading this has heard of this problem and has an idea if there is a solution or if it is hopeless i would be gratefull to heard of it!