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Count cells containing value pairs

    Question

  • This must be harder than I am trying to make it... I have a spreadsheet that is 700 rows long where I need to count each row that contains "10" in column C and the letter "Y" in column F. So far I have tried several variations of Sum and Count and gotten a whole lot of nowhere. Hoping someone can help me out here. I need to do this for a whole lot more than just this example but once I know how to do that its a simple matter of modifying the values in the formula.
    Thursday, June 21, 2012 12:08 AM

Answers

  • On Thu, 21 Jun 2012 00:08:01 +0000, C-M wrote:
     
    >This must be harder than I am trying to make it... I have a spreadsheet that is 700 rows long where I need to count each row that contains "10" in column C and the letter "Y" in column F. So far I have tried several variations of Sum and Count and gotten a whole lot of nowhere. Hoping someone can help me out here. I need to do this for a whole lot more than just this example but once I know how to do that its a simple matter of modifying the values in the formula.
     
    =SUMPRODUCT((C1:C700=10)*(F1:F700="Y"))
     
    or
     
    =COUNTIFS(C1:C700,10,F1:F700,"Y")
     

    Ron
    • Marked as answer by C-M Thursday, June 21, 2012 4:22 PM
    Thursday, June 21, 2012 12:36 AM

All replies

  • On Thu, 21 Jun 2012 00:08:01 +0000, C-M wrote:
     
    >This must be harder than I am trying to make it... I have a spreadsheet that is 700 rows long where I need to count each row that contains "10" in column C and the letter "Y" in column F. So far I have tried several variations of Sum and Count and gotten a whole lot of nowhere. Hoping someone can help me out here. I need to do this for a whole lot more than just this example but once I know how to do that its a simple matter of modifying the values in the formula.
     
    =SUMPRODUCT((C1:C700=10)*(F1:F700="Y"))
     
    or
     
    =COUNTIFS(C1:C700,10,F1:F700,"Y")
     

    Ron
    • Marked as answer by C-M Thursday, June 21, 2012 4:22 PM
    Thursday, June 21, 2012 12:36 AM
  • Thank you! The 2nd answer is exactly what I was looking for. I was not aware of the COUNTIFS
    Thursday, June 21, 2012 4:22 PM