#### Random number Generator

```Could you kindly assist me with a formula that generate 6 numbers between 1
and 49, without repeating any of the integers.
Thanks!
```
 0
RayT (9)
1/14/2008 10:47:00 PM
excel.newusers 15348 articles. 2 followers.

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```I don't believe there is anyway to do this with a formula; you'll need some
VBA code. See http://www.cpearson.com/Excel/randomNumbers.aspx for quite a
few choices. Look at the section entitled "Getting An Array Of Unique,
Non-Duplicated Value" and the UniqueRandomLongs function.

--
Cordially,
Chip Pearson
Microsoft Most Valuable Professional
Excel Product Group, 1998 - 2008
Pearson Software Consulting, LLC
www.cpearson.com
(email on web site)

"RayT" <RayT@discussions.microsoft.com> wrote in message
news:4D3EEAFB-2E4A-4F3E-A59B-60C3DDF1A08F@microsoft.com...
> Could you kindly assist me with a formula that generate 6 numbers between
> 1
> and 49, without repeating any of the integers.
> Thanks!

```
 0
chip1 (1821)
1/14/2008 11:01:44 PM
```Thanks, for takin the time. I will try that? Cheers!

"Chip Pearson" wrote:

> I don't believe there is anyway to do this with a formula; you'll need some
> VBA code. See http://www.cpearson.com/Excel/randomNumbers.aspx for quite a
> few choices. Look at the section entitled "Getting An Array Of Unique,
> Non-Duplicated Value" and the UniqueRandomLongs function.
>
>
> --
> Cordially,
> Chip Pearson
> Microsoft Most Valuable Professional
>     Excel Product Group, 1998 - 2008
> Pearson Software Consulting, LLC
> www.cpearson.com
> (email on web site)
>
> "RayT" <RayT@discussions.microsoft.com> wrote in message
> news:4D3EEAFB-2E4A-4F3E-A59B-60C3DDF1A08F@microsoft.com...
> > Could you kindly assist me with a formula that generate 6 numbers between
> > 1
> > and 49, without repeating any of the integers.
> > Thanks!
>
```
 0
RayT (9)
1/14/2008 11:45:01 PM
```This should work for you:

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.

Each time you  hit <F9>, you'll get a new random selection.

You might choose to place your calc mode into Manual, so that you don't
inadvertently refresh the list.
--
HTH,

RD

---------------------------------------------------------------------------
Please keep all correspondence within the NewsGroup, so all may benefit !
---------------------------------------------------------------------------

"RayT" <RayT@discussions.microsoft.com> wrote in message
news:4D3EEAFB-2E4A-4F3E-A59B-60C3DDF1A08F@microsoft.com...
> Could you kindly assist me with a formula that generate 6 numbers between
> 1
> and 49, without repeating any of the integers.
> Thanks!

```
 0
ragdyer1 (4060)
1/14/2008 11:53:19 PM
```One other play using formulas which might interest you ..

Illustrated in this "Ready-to-randomize" sample:
Randomize 1 - 49 into a 6 col x 8 row grid.xls

The set-up:
Numbers 1 - 49 to be randomized listed in A1:A49
In B1: =RAND()
In C1: =INDEX(\$A:\$A,RANK(B1,B\$1:B\$49))
Copy B1:C1 down to C49

Place in say, E2:
=INDEX(\$C:\$C,ROWS(\$1:1)*6-6+COLUMNS(\$A:A))
Copy E2 across to J2, fill down to J10. Clear F10:J10.
The grid E2:J9 returns 8 sets of 6 random numbers
from the source in col A. E10 returns the last element.

Press F9 to regenerate.
--
Max
Singapore
http://savefile.com/projects/236895
xdemechanik
---
```
 0
demechanik (4694)
1/14/2008 11:55:01 PM
```Hey Max thats some good stuff there man. Its almost the same as how RagDyer
would do, well explained, u make it sound easy. Thanks mate.

"Max" wrote:

> One other play using formulas which might interest you ..
>
> Illustrated in this "Ready-to-randomize" sample:
> Randomize 1 - 49 into a 6 col x 8 row grid.xls
>
> The set-up:
> Numbers 1 - 49 to be randomized listed in A1:A49
> In B1: =RAND()
> In C1: =INDEX(\$A:\$A,RANK(B1,B\$1:B\$49))
> Copy B1:C1 down to C49
>
> Place in say, E2:
> =INDEX(\$C:\$C,ROWS(\$1:1)*6-6+COLUMNS(\$A:A))
> Copy E2 across to J2, fill down to J10. Clear F10:J10.
> The grid E2:J9 returns 8 sets of 6 random numbers
> from the source in col A. E10 returns the last element.
>
> Press F9 to regenerate.
> --
> Max
> Singapore
> http://savefile.com/projects/236895
> xdemechanik
> ---
```
 0
RayT (9)
1/15/2008 12:24:00 AM
```RagDyer, that was again well said, it seems quite simply, i guess if you know
your stuff. It liked it. Thank u all.

"RagDyer" wrote:

> This should work for you:
>
> 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.
>
> Each time you  hit <F9>, you'll get a new random selection.
>
> You might choose to place your calc mode into Manual, so that you don't
> inadvertently refresh the list.
> --
> HTH,
>
> RD
>
> ---------------------------------------------------------------------------
> Please keep all correspondence within the NewsGroup, so all may benefit !
> ---------------------------------------------------------------------------
>
> "RayT" <RayT@discussions.microsoft.com> wrote in message
> news:4D3EEAFB-2E4A-4F3E-A59B-60C3DDF1A08F@microsoft.com...
> > Could you kindly assist me with a formula that generate 6 numbers between
> > 1
> > and 49, without repeating any of the integers.
> > Thanks!
>
>
>
```
 0
RayT (9)
1/15/2008 12:27:00 AM
```Using VBA is very interesting, i really would love to learn alot about that.
Enjoyed working through the steps in the example from the page you mentioned.
I sure would recommend anyone who loves to learn more to try it.  I
personally enjoyed it. Guez there are many ways to skin a cat. Cheers!

"RayT" wrote:

> Thanks, for takin the time. I will try that? Cheers!
>
> "Chip Pearson" wrote:
>
> > I don't believe there is anyway to do this with a formula; you'll need some
> > VBA code. See http://www.cpearson.com/Excel/randomNumbers.aspx for quite a
> > few choices. Look at the section entitled "Getting An Array Of Unique,
> > Non-Duplicated Value" and the UniqueRandomLongs function.
> >
> >
> > --
> > Cordially,
> > Chip Pearson
> > Microsoft Most Valuable Professional
> >     Excel Product Group, 1998 - 2008
> > Pearson Software Consulting, LLC
> > www.cpearson.com
> > (email on web site)
> >
> > "RayT" <RayT@discussions.microsoft.com> wrote in message
> > news:4D3EEAFB-2E4A-4F3E-A59B-60C3DDF1A08F@microsoft.com...
> > > Could you kindly assist me with a formula that generate 6 numbers between
> > > 1
> > > and 49, without repeating any of the integers.
> > > Thanks!
> >
```
 0
RayT (9)
1/15/2008 12:31:02 AM
```welcome, RayT
--
Max
Singapore
http://savefile.com/projects/236895
xdemechanik
---
"RayT" <RayT@discussions.microsoft.com> wrote in message
news:2A74E96D-5B67-43A1-8F98-091EBE376DB4@microsoft.com...
> Hey Max thats some good stuff there man. Its almost the same as how
> RagDyer
> would do, well explained, u make it sound easy. Thanks mate.

```
 0
demechanik (4694)
1/15/2008 12:59:43 AM
```Thanks for your feed-back.
--
Regards,

RD

---------------------------------------------------------------------------
Please keep all correspondence within the NewsGroup, so all may benefit !
---------------------------------------------------------------------------
"RayT" <RayT@discussions.microsoft.com> wrote in message
news:F397EA17-27D8-4314-A4A1-EC48972953BD@microsoft.com...
> RagDyer, that was again well said, it seems quite simply, i guess if you
> know
> your stuff. It liked it. Thank u all.
>
> "RagDyer" wrote:
>
>> This should work for you:
>>
>> 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.
>>
>> Each time you  hit <F9>, you'll get a new random selection.
>>
>> You might choose to place your calc mode into Manual, so that you don't
>> inadvertently refresh the list.
>> --
>> HTH,
>>
>> RD
>>
>> ---------------------------------------------------------------------------
>> Please keep all correspondence within the NewsGroup, so all may benefit !
>> ---------------------------------------------------------------------------
>>
>> "RayT" <RayT@discussions.microsoft.com> wrote in message
>> news:4D3EEAFB-2E4A-4F3E-A59B-60C3DDF1A08F@microsoft.com...
>> > Could you kindly assist me with a formula that generate 6 numbers
>> > between
>> > 1
>> > and 49, without repeating any of the integers.
>> > Thanks!
>>
>>
>>

```
 0
ragdyer1 (4060)
1/15/2008 1:04:02 AM
```On Jan 14, 3:53=A0pm, "RagDyer" <ragd...@cutoutmsn.com> wrote:
> In an out-of-the-way location, say Column Z, enter the Rand
> function[.]  In Z1 enter =3DRand()[.]  And copy down to Z49.
>
> Then enter this formula wherever you wish:
> =3DINDEX(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 =3DRANK(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.
```
 0
1/15/2008 1:28:07 AM
```On Jan 14, 3:53=A0pm, "RagDyer" <ragd...@cutoutmsn.com> wrote:
> In an out-of-the-way location, say Column Z, enter the Rand function[.]
> In Z1 enter =3DRand()[.]  And copy down to Z49.
>
> Then enter this formula wherever you wish:
> =3DINDEX(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 =3DRANK(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.
```
 0
1/15/2008 5:09:37 AM
```On Jan 14, 9:09 pm, I wrote:
> On Jan 14, 3:53 pm, "RagDyer" <ragd...@cutoutmsn.com> 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.
> [....]
> 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.

Only out of context.  ROW(\$A\$1:\$A\$49) returns the array {1,2,...,49}.
In the context of INDEX(), the RANK() result (1,2,...,49) is used to
index into that array.

In this context, I believe that using INDEX() and ROW() is redundant,
since the OP is interested in randomly choosing amount 1,2,...,49,
which is exactly what RANK() returns, given that Z1:Z49 contains
random values.

However, if the OP had been interested in, for example, randomly
selecting from the range 13-61, then perhaps INDEX(ROW(\$A\$13:\$A
\$61),RANK(Z1,\$Z\$1:\$Z\$49)) could be used -- although, I believe that
12+RANK(Z1,\$Z\$1:\$Z\$49) would suffice.

On the other hand, if the OP had wanted to randomly select unique
values from A1:A49, then I believe INDEX(\$A\$1:\$A\$49,RANK(Z1,\$Z\$1:\$Z
\$49)) could be used, copying down for as many selections as required.

RagDyer (or any other expert), please comment.
```
 0
1/15/2008 4:43:09 PM
```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
-----------------------------------------------------------------------------------------------
Please keep all correspondence within the Group, so all may benefit !
-----------------------------------------------------------------------------------------------

<curiousgeorge408@hotmail.com> wrote in message
On Jan 14, 3:53 pm, "RagDyer" <ragd...@cutoutmsn.com> 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.

```
 0
ragdyer1 (4060)
1/15/2008 5:02:28 PM
```Sorry, just detected that the earlier sample contained an inadvertent error*

Here's the corrected version:
Randomize 1 - 49 into a 6 col x 8 row grid.xls

*The formulas in the output grid E2:J10 got messed up. To correct, just
re-copy E2 across/down to J10. Clear F10:J10. The grid E2:J9 will return 8
sets of 6 random numbers from the source in col A. E10 returns the last
element.
--
Max
Singapore
http://savefile.com/projects/236895
xdemechanik
---

```
 0
demechanik (4694)
1/15/2008 11:16:02 PM

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I have 2003 version of Word at work. I am trying to create an outline numbered list with 3 levels. When I hit the enter button after typing the text for an item, word does not seem to recognize that it should continue the list. The next line just has normal formatting, and I'm forced to use the format painter brush to copy a previous item from the list, and then Word seems to recognize it as part of the list. I have verified the "Apply Automatic Numbered Lists" is checked in the Tools/Autocorrect Options/Autoformat as you type menu. An observation about Word ...

How do I prevent duplicate numbers in a range of validated cells?
I have a validation rule for a series of cells. The 3 cells are only allowed to have a value of 3, 5, or 1. I want to fix it so that each number may only be used once in a range of 3 cells. .... So in A1:A3, each number may only appear once or an error message pops up or the cell starts blinking ~ something needs to happen to notify the user that there is an error that needs correcting. THanks ...

Random numbers
How exactly do i generate random numbers in VC++.Is the a function similar to random() in C++.Thanks in advance Fenn On Mon, 10 Jan 2005 22:42:52 -0800, "Fenn" <fenn_j@yahoo.com> wrote: >How exactly do i generate random numbers in VC++.Is the a function similar >to random() in C++.Thanks in advance > VC++i is C++. Besides, there is no random() in C++. I think you mean rand(). And besides, that is a C function imported into C++. In any event, generate random numbers in Visual C++ using MFC exactly the same way you would in any C++ (or C) program. .. yeah, b...

debugger shows wrong line number
I am experiencing a rather strange behavior -- when I step through my program, the debugger shows the debugging cursor at the wrong line. Also, when I set a breakpoint at any line, the debugger stops at a different line (sometimes the next line or the one after). Any idea why I might be seeing this problem. I am using Visual Studio .NET 2003 >I am experiencing a rather strange behavior -- when I step through my >program, the debugger shows the debugging cursor at the wrong line. > >Also, when I set a breakpoint at any line, the debugger stops at a different >line (someti...