Consider the expansion of . Then the sum of all the coefficients of the terms
Which contains all of a, b, c and d is
We have
Sum of coefficients which contains all of a, b, c and d
= Number of ways of distributing six distinct objects in four boxes such that no box remains empty
Which contains a but not b is
Sum of coefficients which contains a but not b
= Number of ways of distributing six distinct objects in three boxes (a, c, d)
- Number of ways of distributing six distinct objects in two boxes (c, d)
= 36 - 26 = 665
Which contains both a and b is
Sum of coefficients which contain both a and, b
= Number of ways of distributing six distinct objects in four boxes (a, b, c, d)
- Number of ways of distributing six distinct objects in three boxes (a, c, d)
- Number of ways of distributing six distinct objects in three boxes (b, c, d)
+ Number of ways of distributing six distinct objects in two boxes (c, d)