Hello RagDyeR from Steved

I am using your formula and so far I have'nt got any doubles.

I've tried it several times yet to produce a double.

So yes I'm Using it.

Thankyou

"RagDyeR" wrote:

You're *not* looking for random numbers, since true randomness accepts the
possibility of duplication, since each item has an equal chance of occurring
at each interval.

You're looking for a random *order* of specific items (numbers 1 to 40).

So, you therefore need, first of all, the specific items so that they can be
rearranged randomly.
That's the numbers of the rows, 1 to 40, making the construction of an
actual list unnecessary.

Next, the list is rearranged virtually, by accessing the 40 columns of
random numbers and ranking these numbers, where this ranking is mirrored in
the indexed row numbers.
Even if these *true* random numbers are duplicated, and they are ranked
*equally*, since they designate the row numbers, there *cannot* be a
duplicated return, since there are *no* duplicate row numbers existing.

And since each row must be independent of the other rows, so that
duplication is avoided, you need a separate row of random numbers to be
ranked differently from the others.

Therefore, my hat is off to Harlan if he can accomplish this scenario
*without* the existence of this 40 column by 10 row array of random numbers.

--

Regards,

RD
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Please keep all correspondence within the Group, so all may benefit !
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"Steved" wrote in message
...
Hello Harlan from Steved

Harlan done as discribed below but still have the issue offsame number twice
omn the same line

in this case row 8 i've 13 twice and row 10 I have 24 twice.

Is it possible please to have this issue where their are no doubles.
Thankyou for your patience.

"Harlan Grove" wrote:

"Steved" wrote...
....
=COUNTIF(PRNA,"<="&TRANSPOSE(OFFSET(PRNA,INT(35*R AND()),0,6,1)))
....

Sorry, I screwed this up. Swap the 6 and 1 arguments, so

=COUNTIF(PRNA,"<="&TRANSPOSE(OFFSET(PRNA,INT(35*RA ND()),0,1,6)))