The number of rational terms in the expansion of (1+2+33)6, is

We have,(1+2+33)6=∑r+s+t=6 6!r!s!t!×2s/23t/3Now, 6!r!s!t!×2s/2×3t/3 will be an integer, if s is a multiple of2 and t is a multiple of 3 i.e..s=0, 2, 4, 6 and  t=0, 3, 6But, r+s+t=6  Therefore, we have the following possibilities.r=6,s=0,t=0r=3,s=0,t=3r=0,s=0,t=6r=4,s=2,t=0r=1,s=2,t=3r=2,s=4,t=0r=0,s=6,t=0Hence, there are 7 rational terms in the expansion of(1+2+33)6