4-seed K-State going dancing in San Jose against UC-Irvine

Kansas State guard Mike McGuirl (00) and forward Austin Trice (23) celebrates with teammates during the second half of an NCAA college basketball game against TCU in the quarterfinals of the Big 12 conference tournament in Kansas City, Mo., Thursday, March 14, 2019. (AP Photo/Orlin Wagner)
By  | 

(WIBW) -- It's official. No. 15 Kansas State is going dancing for a third-straight year.

The Wildcats will head into the San Jose regional as the 4-seed with a first-round matchup against UC-Irvine slated for Friday, March 22, at 1 p.m. CT in San Jose. That game can be watched on TBS.

CLICK HERE TO GET YOUR PRINTABLE GRAPHIC

The four-slot is K-State's best seeding since 2013, the last time the Wildcats won the Big 12 regular season.

Virginia landed the No. 1 seed in the South. Should KSU win the first round, it would face either five-seed Wisconsin or 12-seed Oregon in the second round on Sunday.

The South region's semifinals and finals will be played in Louisville, Ky.

This is the fifth time in head coach Bruce Weber's seven years as K-State's head coach that the Wildcats will appear in the Big Dance.

Kansas State finished the regular season 24-7 after sharing a piece of the Big XII championship with Texas Tech and going 14-4 in conference play.

Senior forward Dean Wade is battling a foot injury and is projected as doubtful for the NCAA tournament--which could mean he'll miss his second-straight.

UC-Irvine went 30-5 overall and 15-1 in Big West conference play. The Anteaters will enter Friday's first-round game on a 16-game winning streak.

The two team's last matchup was in November of 2017, and resulted in the Wildcats winning 71-49 in Bramlage Coliseum.

K-State is joined by Oklahoma as the only other Big 12 team in the South Region. The Sooners are the No. 9 seed.