Question 128 of 278EasyNUMERICAL
Number of integral solutions to the equation x+y+z=21, where x≥1,y≥3,z≥4, is equal to \_\_\_\_\_\_\_.
Answer: 105 Solutions · 1
Official solution
Substituting x′=x−1≥0, y′=y−3≥0, z′=z−4≥0: x′+y′+z′=21−8=13.
\[\binom{13+3-1}{3-1} = \binom{15}{2} = 105\] Answer the question to unlock