Excel formula challenge: Words with all vowels (MMULT)

Excel formula challenge Vowels: A video demonstrating the formula breakdown to an Excel challenge on getting words containing all vowels. It showcases the use of the MMULT function to get his working

Link to Excel Challenge:

Link to Excel workbook:

[IvanCortinas_ES]

Before all of you, one of the best Youtube channels and one of the best professionals. Thank you Victor.

[rtrbs]

Thanks Victor FILTER(A903:A911, BYROW(ISNUMBER(SEARCH(C902:G902, A903:A911)), LAMBDA(a, AND(a))))

[petercompton]

I have never used MMULT before - thanks for that. Another great video

[wayneedmondson]

Interesting use of MMULT! My first pass was with BYROW, as follows: =LET(rng, A2:A10, FILTER(rng, BYROW(--ISNUMBER(SEARCH({"a", "e", "i", "o", "u"}, rng)), LAMBDA(r, SUM(r)))=5)). Will study yours. Thanks always for the challenge and insight. Thumbs up!

[supaflyutubeguy]

Your channel is amazing! Thank you for sharing.

[quadriatharutech]

I gained something new. I enjoyed the mathematical approach of raising the sequence by Zero. I could have thought of doing SEQUENCE (5)/SEQUENCE (5). This is awesome. I need to learn more about the MMULT function. Thanks Mr. Victor

[williamarthur]

That was really good, I must get to grips with using MMULT, I know it's been around for ages, but as someone who's a bit shaky with dot multiplication I've tended to avoid it, thanks for the inspiration,
you'll be gaining anew subscriber.

[jazzista]

Victor. Happy late new year my friend. Excellent explanation of the MMULT function. I also liked your trick of raising the sequence result by the zero power to force an array of 1s. Pretty elegant

[mohamedsoffar]

Thanks for your amazing content my dear >>>

instead of using MMULT we can use BYROW function 😍

3:27
you write BYROW and i thought that you are going to use 🙃

=BYROW(YourArray, LAMBDA(x, AND(x)))

[ogunronbitaiwofisayo]

Thank you Victor. But instead of the MMULT, why not just count instead.
Something like: =LET(a, A2:A10, FILTER(a, MAP(a, LAMBDA(x, COUNT(SEARCH({"a", "e", "i", "o", "u"}, x))>=5))))

[olukunlebabajide]

Superb as always. Won't PRODUCT work too?