Given,
Sr={(x,y,z):x+y+z=11,x≥r,y≥r,z≥r}
S2={(x,y,z):x+y+z=11,x≥2,x≥2,z≥2}
∵x≥2⇒x−2≥0,y−2≥0,z−2≥0
Let
x−2=a,y−2=b,z−2=c ∴x+y+z=11⇒a+b+c=5,a,b,c≥0
n(S2)=5+3−1C3−1=7C2=21 Similary, for
S3,a+b+c=2 n(S3)=2+3−1C2=4C2=6 For
S4,a+b+c=−1 (not possible)
∴n(S2)+n(S3)+n(S4)=21+6+0=27