In this particular case, you're right, the Rand() function would suffice.

However, I always suggest the combination with Index() as a more or less
*generic* solution, which can be understood and built on.

What if the OP *later* wanted numbers *other* then 1 to 49, say 10 to 58, or
100 to 148?

=INDEX(ROW($A$100:$A$148),RANK(Z1,$Z$1:$Z$49))

OR, say that the list of numbers are *not consecutive*, OR, say that a
random list of names is desired ... with the master list located at say J50
to J98:

=INDEX($J$50:$J$98,RANK(Z1,$Z$1:$Z$49))

So, you can see how easily a revision might be accomplished if the necessary
basic functions are presented at the outset.
--

Regards,

RD
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Please keep all correspondence within the Group, so all may benefit !
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wrote in message
...
On Jan 14, 3:53 pm, "RagDyer" wrote:
In an out-of-the-way location, say Column Z, enter the Rand
function[.] In Z1 enter =Rand()[.] And copy down to Z49.

Then enter this formula wherever you wish:
=INDEX(ROW($A$1:$A$49),RANK(Z1,$Z$1:$Z$49))
Copy down as many rows as you need random numbers.

Can you please explain the theory of operation.

I agree that it works. But it seems that =RANK(Z1,$Z$1:$Z$49) would
suffice. RANK(Z1,$Z$1:$Z$49) seems to return the same thing that the
full INDEX expression returns. It seems that ROW($A$1:$A$49) is
always 1.

In any case, I agree with RayT that this is a clever, yet nicely
simple solution to the problem.