Consider the expansion of (a+b+c+d)6. Then the sum of all the coefficients of the termsWhich contains all of a, b, c and d isWhich contains a but not b isWhich contains both a and b is

We have (a+b+c+d)6Sum 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 =46−4C136+4C226−4C116=1560Sum 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 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)=46−2×36+2C2×26=2702