Question

There are $(a+b+c)$ things in which 'a' things are of same kind, 'b' are of another same kind and rest 'c' things are different kind then the number of ways in which one or more things can be selected is

• A. $(a+1)(b+1)2^c-1$
• B. $2^{a+b+c}-2$
• C. $2^{a+b}(2^c-1)$
• D. $2^{a+b}(2^c+1)-1$

Answer 1

The correct option is A $(a+1)(b+1)2^c-1$

Here, all the cases of selection from $a,b,c$ are independent
in the 'a' things are of same kind, you can select $0,1,2,\mathrm{3....}$or a of them. so there are $(a+1)$ ways to do it
Similarly $(b+1)$ ways of doing it with the b things
in the c dissimilar things, there are 2 choices of being selected or not selected for each things and there are c such things. so, there are $2^c$ ways of doing it
$\therefore$ Number of ways of selecting things$=(a+1)(b+1)2^c$
in the above cases, there exists a case in which nothing is selected
So, number of ways in which one or more things can be selected $=(a+1)(b+1)2^c-1$