Prove PB(8, {4,3}, 1) does not exist.

Question

I was reading Wallis, Intro to Combinatorial Design.

This is 2.1.3. I couldn't understand the way of counting. Hint says, let $f_4, f_3$ be the block counts of sizes 4,3 respectively, then,

$6f_4 + 3f_3 = 28$ a contradiction, 3 divides left but not right.

thanks in advance.

Notation: $PB(v, K, \lambda)$ v: number of vertices, K set of sizes of blocks, lambda: unique pair repetition count.

Answer 1

Figured it out.

${4 \choose 2} = 6$, ${3 \choose 2} = 3$, ${8 \choose 2} = 28$.