Question

There are counters available in 3 different colours. The counters are all alike except for the colour. The total number of arrangement consisting of 5 counters, assuming sufficient number of counters of each colour if each arrangement consists of counters atleast one of each colour is :

• A. ${ }^5 C _3 \cdot 3 ! \cdot 3^2$
• B. $3^5-\left({ }^3 C _2 \cdot 2^5+{ }^3 C _1\right)$
• C. $3\left(3^4-2^5\right)$
• D. ${ }^3 C _1 \cdot \frac{5 !}{3 !}+{ }^3 C _2+\frac{5 !}{2 ! \,\,2!}$

Answer 1

Answer: (D)

A A A ............, B B B ............., C C C .............
$XXXXX ; 3^5-[$ No. of ways when exactly one colour counter is not there + when exactly two colour counters are not there]
$=3^5-\left[{ }^3 C _1\left(2^5-2\right)+{ }^3 C _2\right]$ ......(1)
altemately
A B C X X
we can choose both alike counters for two crosses or we can choose both different.
Accordingly
${ }^3 C _1 \cdot \frac{5 !}{3 !}+{ }^3 C _2 \cdot \frac{5 !}{2 ! 2 !}$ ............(2)
(1) and (2) gives $150 \Rightarrow D$