If one of multiple wildcards appears in 5 different cells, then display value in adjacent cell

[CF64]

Hello and Happy New Year,
I am trying to find a way to look across a range of 6 cells in the same row, to find any of multiple wildcard values and, in an adjacent cell, return a "yes" if any one of the multiple wildcards is found in any of the cells in the range, or "no" if not.

Initially, I was able to use the following formula in column E, when I was just looking in 4 cells...
=IF(SUM(COUNTIF(a2,{"*hockey*","*field*","*baseball*"})+COUNTIF(b2,{"*hockey*","*field*","*baseball*"})+COUNTIF(c2,{"*hockey*","*field*","*baseball*"})+COUNTIF(d2,{"*hockey*","*field*","*baseball*"})),"YES","NO")

...but when I added the 2 additional cells to check (F2 and G2), I received the "you've entered too many arguments"

Also, in the 2 additional cells I would be looking for different wildcard values than the preceding 4 cells.

Any advice is very much appreciated

 

[etaf]

what wild cards are you looking for
Why not use the range
=COUNTIF(A2:D2,"*hockey*")+COUNTIF(A2:D2,"*field*")+COUNTIF(A2:D2, "*baseball*")
Then you can add the other criteria for F2 & G2
rather than an array

=IF(COUNTIF(A2:D2,"*hockey*")+COUNTIF(A2:D2,"*field*")+COUNTIF(A2:D2, "*baseball*")+COUNTIF(F2:G2,"*Tennis*")+COUNTIF(F2:G2, "*football*"),"YES","NO")

Book1
	A	B	C	D	E	F	G	H
1
2						football
3
4
5				A-D	NO	+	F-G	YES
6				Count	0
7
Sheet2

Cell Formulas
Range	Formula
H5	H5	=IF(COUNTIF(A2:D2,"*hockey*")+COUNTIF(A2:D2,"*field*")+COUNTIF(A2:D2, "*baseball*")+COUNTIF(F2:G2,"*Tennis*")+COUNTIF(F2:G2, "*football*"),"YES","NO")
E5	E5	=IF(COUNTIF(A2:D2,"*hockey*")+COUNTIF(A2:D2,"*field*")+COUNTIF(A2:D2, "*baseball*"),"YES","NO")
E6	E6	=COUNTIF(A2:D2,"*hockey*")+COUNTIF(A2:D2,"*field*")+COUNTIF(A2:D2, "*baseball*")

 

[Fluff]

Another option

=IF(SUM(COUNTIF(A2:D2,{"*hockey*","*field*","*baseball*"}))+SUM(COUNTIF(F2:G2,{"*Rugby*","*curling*"})),"YES","NO")

 

[CF64]

Thank you both. Much appreciated.

 

[Fluff]

Glad we could help & thanks for the feedback.