# Words formed using the letters from the word $CHISEL$

How many words of 6 letters can be formed using the letters from the word $"CHISEL"$ such that the words don't contain $"LE"$ OR $"CHEL"$?

I have tried considering $"LE"$ and $"CHEL"$ as two characters but I'm a bit confused.

• Please clarify whether repetition of letters is allowed. For example, is "CHCIEL" allowed? Or must every letter be used once in each 6 letter word? – Namaste Mar 2 '14 at 17:28
• Repetition is not allowed. – Lstoi Mar 2 '14 at 17:50
• Good, then my answer applies! – Namaste Mar 2 '14 at 17:51

I. First calculate the number of words that can be formed from "CHISEL", with no restrictions.

Subtract from your answer to $I$ the number of words that contain (the "chunk" "LE" or the chunk "CHEL").

Assuming repetition of letters is not allowed, then we we have $6$ possible letters to choose from for the first letter of the word, $5$ possible choices for the second letter...etc. In all, there are $6\times 5\times 4\times 3\times 2\times 1 = 6!$ possible words with no restrictions.

(II.) Since we are assuming no repetition, it will never be the case that "LE" and "CHEL" appear in the same word, since clearly $L$ and $E$ would be repeating.

• The number of words containing "CHEL": Taken as a "chunk" (consider the letters glued together); then there are three choices where to place "CHEL".

$$CHEL\;X\;X,\quad X\;CHEL\;X,\quad X\;X\;CHEL.$$ Then there are $2$ choices remaining for where to place "I", and the position of the remaining letter is then determined. Total "CHEL"-containg words: $3\times 2 = 6$.

• Use the same procedure to determine the number of possible words containing "LE". You should find that there are $5$ places in which the chunk "LE" can be placed, with $4!$ choices for where the other $4$ letters are placed: Total "LE"-containing words: $5 \times 4! = 5!$.

Now, we subtract each result in (II.) from the result for (I.): $$6! -(6 + 5!) = 720 - (6 + 120) = 594 \;\text{possible words}$$

Find the total number of words that can be formed using the letters C-H-I-S-E-L, and subtract the number of words containing LE or CHEL, noting that no word can contain both.

In order to find the number of words containing, for example, CHEL, we may consider CHEL as its own character, so that $$n_{CHEL} = 3 \times 2 \times 1$$ since there are $3$ characters total to move around, under this restriction. Similarly, $$n_{LE} = 5 \times 4 \times 3 \times 2 \times 1$$ So, in total, the number of suitable arrangements is $$n = n_{TOTAL} - n_{CHEL} - n_{LE}$$

Hint: Try counting how many words contain $LE$, and how many words contain $CHEL$. Then subtract the sum of those two from the total number of possible words.