Question

The number of ways in which 6 different nuts can be used in machine A, B, C so that no machine is left without the nut is:-
A) 540
B) 180
C) 270
D) None of the above

Answer 1

Hint: The problem can be solved by breaking them into sub-problem and then solving them separately. So consider all the different possibilities of processing nuts in these 3 machines, without leaving any machine empty. Also, consider that both the machines and the nuts are different. hence, they are also rearranged among themselves.

Complete step by step Answer:

The total number of nuts 6
Total number of machine 3 (A, B, C)
Since no machine should be left empty therefore each machine should have at least one nut. Now since there are three machines are so the possible cases are:
case 1: Two machines have 1 nut, one machine has 4 nuts.
M1=1 M2 = 1 M3,= 4
The possible number of ways in which we can choose this combination is
For one nut 6C1
Since one net is already taken from 6, there are only 5 left.
Now no. of ways we can choose one nut out of 5 = 5C1
Now again 1nut is taken out of 5, so the left nuts are 4
Now no of ways to choose 4 nuts out of 4 nuts is = 4C4
So total number of ways to select the nuts as given in the above case$={}^{6}{{C}_{1}}\ \times \ {}^{5}{{C}_{1}}\ \times \ {}^{4}{{C}_{4}}\ \times \ \dfrac{3!}{2!}$
\[\text{=6}\ \times \ \text{5}\times \text{1}\times \dfrac{\text{6}}{2}\]
\[\text{=}\ \text{90}\]

We multiplied an extra term i.e. \[\dfrac{\text{3!}}{\text{2}!}\] because the nuts will rearrange among 3 different machines and since two machines will have the same no. of nuts.
case 2: The machine will have combination of nuts as
                  M1 = 1 M2 = 2 M3,= 3
Now as explained above
The total number of ways to select nuts\[={}^{\text{6}}{{\text{C}}_{\text{1}}}{}^{\text{5}}{{\text{C}}_{\text{2}}}{}^{\text{3}}{{\text{C}}_{\text{3}}}\ \times \ \text{3!}\]
$=6\ \times \ \dfrac{5!}{3!2!}\ \times \ 1\times \ 3!$
$=6\ \times \ 5\ \times \ 4\ \times \ 3$
$=120\ \times \ 3$
$=360$


Here each machine has a different number of nuts, therefore there are 3! ways to rearrange the machine.
Case 3: Each machine have 2 nuts
M1 = 2 M2 = 2 M3 ,= 2

The total number of possible ways to select nuts \[={}^{6}{{C}_{2}}\ \times \ {}^{4}{{C}_{2}}\ \times \ {}^{2}{{C}_{2}}\ \times \ \dfrac{3!}{3!}\]
$=\dfrac{6!}{2!4!}\ \times \ \dfrac{4!}{2!2!}\ \times \ \dfrac{2!}{2!0!}\ \times \ \dfrac{3!}{3!}$
$=\dfrac{6!}{2!}\times \ \dfrac{1!}{2!}\times \ \dfrac{1!}{2!}$
$=\dfrac{6\ \times \ 120}{8}$
$=3\times 30$
$=90$
Now total number of ways in which 6 nuts can be placed in three different machine so that no machine is empty sum of all possible cases, i.e. (case 1 case 2 case 3)
\[\text{=}\ \text{90}\ \text{+}\ 3\text{60}\ \text{+ 90}\]
$=180\ +\ 360$
$=\ 540$
So the correct option is A

Note: Always consider all the different possible cases so that you do not miss out any possible way. Additionally, remember that if the machines were identical then we will not consider their rearrangement.