#### How can I count a column of numbers

```and return the number of values that fall within certain criteria?

For example, if I have 300 salaries listed, and wanted to group them into
these ranges:
3000-5999
6000-8999
9000-11999
12000-14999

etc.

Countif (as far as I know) can only have one argument.

TIA!

```
 0
bob84 (4)
10/24/2003 1:51:56 AM
excel.misc 78881 articles. 5 followers.

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```You may want to look at =frequency() in help.

Bob wrote:
>
> and return the number of values that fall within certain criteria?
>
> For example, if I have 300 salaries listed, and wanted to group them into
> these ranges:
> 3000-5999
> 6000-8999
> 9000-11999
> 12000-14999
>
> etc.
>
> Countif (as far as I know) can only have one argument.
>
> TIA!

--

Dave Peterson
ec35720@msn.com
```
 0
ec35720 (10082)
10/24/2003 2:03:42 AM
```Actually, you can still use countif

=COUNTIF(A:A,">=3000")-COUNTIF(A:A,">5999")

for 3000-5999

=COUNTIF(A:A,">=6000")-COUNTIF(A:A,">8999")

and so on

you might want to replace these hard coded values with cell references and
change
the criteria there.

It can also be done using sumproduct

=SUMPRODUCT((A1:A1000>=6000)*(A1:A1000<=8999))

--

Regards,

Peo Sjoblom

" Bob" <bob@bobsworld.com> wrote in message
news:0B%lb.125316\$qj6.7804315@news1.news.adelphia.net...
> and return the number of values that fall within certain criteria?
>
> For example, if I have 300 salaries listed, and wanted to group them into
> these ranges:
> 3000-5999
> 6000-8999
> 9000-11999
> 12000-14999
>
> etc.
>
> Countif (as far as I know) can only have one argument.
>
> TIA!
>
>

```
 0
terre081 (3244)
10/24/2003 2:03:53 AM
```Thanks guys!  Those work, and this did also:

{=SUM(IF(A2:A475<4000,1,0))}
{=SUM(IF(A2:A475>3999,IF(A2:A475<8001,1,0)))}

etc...

"Peo Sjoblom" <terre08@mvps.org> wrote in message
news:O2xfVMdmDHA.988@TK2MSFTNGP10.phx.gbl...
> Actually, you can still use countif
>
> =COUNTIF(A:A,">=3000")-COUNTIF(A:A,">5999")
>
> for 3000-5999
>
> =COUNTIF(A:A,">=6000")-COUNTIF(A:A,">8999")
>
> and so on
>
> you might want to replace these hard coded values with cell references and
> change
> the criteria there.
>
> It can also be done using sumproduct
>
>
> =SUMPRODUCT((A1:A1000>=6000)*(A1:A1000<=8999))
>
> --
>
> Regards,
>
> Peo Sjoblom
>
> " Bob" <bob@bobsworld.com> wrote in message
> news:0B%lb.125316\$qj6.7804315@news1.news.adelphia.net...
> > and return the number of values that fall within certain criteria?
> >
> > For example, if I have 300 salaries listed, and wanted to group them
into
> > these ranges:
> > 3000-5999
> > 6000-8999
> > 9000-11999
> > 12000-14999
> >
> > etc.
> >
> > Countif (as far as I know) can only have one argument.
> >
> > TIA!
> >
> >
>
>

```
 0
bob84 (4)
10/24/2003 3:13:47 AM
```Yes, the array formula will work, but it isn't needed. In general you should use array formulas
when there is no other solution.

On Fri, 24 Oct 2003 03:13:47 GMT, "                              Bob" <bob@bobsworld.com> wrote:

>Thanks guys!  Those work, and this did also:
>
>{=SUM(IF(A2:A475<4000,1,0))}
>{=SUM(IF(A2:A475>3999,IF(A2:A475<8001,1,0)))}
>
>etc...
>
>"Peo Sjoblom" <terre08@mvps.org> wrote in message
>news:O2xfVMdmDHA.988@TK2MSFTNGP10.phx.gbl...
>> Actually, you can still use countif
>>
>> =COUNTIF(A:A,">=3000")-COUNTIF(A:A,">5999")
>>
>> for 3000-5999
>>
>> =COUNTIF(A:A,">=6000")-COUNTIF(A:A,">8999")
>>
>> and so on
>>
>> you might want to replace these hard coded values with cell references and
>> change
>> the criteria there.
>>
>> It can also be done using sumproduct
>>
>>
>> =SUMPRODUCT((A1:A1000>=6000)*(A1:A1000<=8999))
>>
>> --
>>
>> Regards,
>>
>> Peo Sjoblom
>>
>> " Bob" <bob@bobsworld.com> wrote in message
>> news:0B%lb.125316\$qj6.7804315@news1.news.adelphia.net...
>> > and return the number of values that fall within certain criteria?
>> >
>> > For example, if I have 300 salaries listed, and wanted to group them
>into
>> > these ranges:
>> > 3000-5999
>> > 6000-8999
>> > 9000-11999
>> > 12000-14999
>> >
>> > etc.
>> >
>> > Countif (as far as I know) can only have one argument.
>> >
>> > TIA!
>> >
>> >
>>
>>
>

```
 0
myrnailarson (145)
10/24/2003 3:57:43 AM
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