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Question: The number of four-letter words that can be formed with letters a, b, c such that all three letters ...

The number of four-letter words that can be formed with letters a, b, c such that all three letters occur is
A.30
B.36
C.81
D.256

Explanation

Solution

Hint- In order to solve this question we will make the sets of possible selections and by using the basic concept of permutation. A permutation is an arrangement of all or part of a set of objects, with regard to the order of arrangement. We will use this definition to reach the answer.

Complete step-by-step answer:
The 4 letter code will have a, b, c and a repeat letter from either a, b or c.
The possible selections are
\left\\{ {a,a,b,c} \right\\},\left\\{ {b,b,a,c} \right\\},\left\\{ {c,c,a,b} \right\\}
First selection is \left\\{ {a,a,b,c} \right\\} = \dfrac{{4!}}{{2!}}
=4×3×22=12= \dfrac{{4 \times 3 \times 2}}{2} = 12
Second selection is \left\\{ {b,b,a,c} \right\\} = \dfrac{{4!}}{{2!}}
=4×3×22=12= \dfrac{{4 \times 3 \times 2}}{2} = 12
Third selection is \left\\{ {c,c,a,b} \right\\} = \dfrac{{4!}}{{2!}}
=4×3×22=12= \dfrac{{4 \times 3 \times 2}}{2} = 12
Therefore number of four letter words that can be formed with letters a, b, c such that all three letters occur is =12+12+12=36 = 12 + 12 + 12 = 36
Hence, the correct option is “B”.

Note- In order to solve these types of questions, learn the concept of permutation and combination. The above question can be also solved by using the fundamental principle of counting which states that if a thing can be arranged in m ways and other in n ways then the total number of ways a thing can be arranged in m×nm \times n ways.