Question

The number of ways in which 6 different nuts can be used in machines $A,B,C$ so that no machine is left with out the nut is

• A. $540$
• B. $180$
• C. $270$
• D. None of these

Answer 1

The correct option is A $540$

Number of nuts$=6$, Number of machines$=3$
Number of case in which nuts can be used in machines $A,B,C$ are

1st	2nd	3rd
ABC	ABC	ABC
114	123	222

Now number of ways for case$I$ $={}^6{C}_1{\times }^5{C}_1{\times }^4{C}_4\times \frac{3!}{2!}=90$
Number of ways for case$II$ $(1,2,3)$ $={}^6{C}_1{\times }^5{C}_2{\times }^3{C}_3\times \frac{3!}{1!1!1!}=360$
Number of ways for case$III$ $(2,2,2)$ $={}^6{C}_2{\times }^4{C}_2{\times }^2{C}_2\times \frac{3!}{3!}=90$
$\therefore$ Total number of ways $=90+360+90=540$