Question

The letters of the word ‘ZENITH’ are written in all possible orders. How many words are possible if all those words are written out as in a dictionary? What is the rank of the word ‘ZENITH’?

Answer 1

Hint: Use the following steps to reach the solution:
Step:1 Write down the letters in alphabetical order.
Step:2 Find the number of words that start with a superior letter.
Step:3 Solve the same problem, without considering the first letter.
So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:

In a dictionary the words at each stage are arranged in alphabetical order. In the given problem we must therefore consider the words beginning with E, H, I, N, T, Z in order.

So, first we find out how many words are formed with each letter of the word ZENITH

Number of words starting with E = 5! = \[5 \times 4 \times 3 \times 2 \times 1\]= 120

Number of words starting with H = 5! = \[5 \times 4 \times 3 \times 2 \times 1\]= 120

Number of words starting with I = 5! = \[5 \times 4 \times 3 \times 2 \times 1\]= 120

Number of words starting with N = 5! = \[5 \times 4 \times 3 \times 2 \times 1\]= 120

Number of words starting with T = 5! = \[5 \times 4 \times 3 \times 2 \times 1\]= 120

Number of words starting with Z = 5! = \[5 \times 4 \times 3 \times 2 \times 1\]= 120

But one of these words will be the word ZENITH itself.

So, we first find the word begin with ZEH, ZEI & ZENH

Number of words starting with ZEH = 3! = \[3 \times 2 \times 1\] = 6

Number of words starting with ZEI = 3! = \[3 \times 2 \times 1\] = 6

Number of words starting with ZENH = 2! = \[2 \times 1\] = 2

So, we found the first word beginning with ZENI is the word ZENIHT and the next word is ZENITH. Thus, here 2 more words will be counted.

Therefore, the rank of the word ZENITH = 120+120+120+120+120+6+6+2+2 = 616

Thus, the rank of the word ZENITH is 616.

Note: A common type of problem in many examinations is to find the rank of a given word in a dictionary. What this means is that you are supposed to find the position of that word when all permutations of the word are written in alphabetical order. Of course, the words do not need to have any meaning. Since there are \[n!\] different words that are possible, a few simple tricks can be used to minimise the effort needed.