$\binom{n}{3}\geq 12$
$\frac{n!}{(n-3)!*3!}\geq 12$
$\frac{n*(n-1)*(n-2)}{6}\geq 12$
$n*(n-1)*(n-2)\geq 72$
Considering $n$$=5$ we get $5*4*3=60$$\ngeqslant 72$
Now considering $n=6$ we get $6*5*4= 120$$> 72$.
$\therefore$ Minimum number of jokes needed $=$ $6$.