Question

In how many different ways can the letters of the word ‘SALOON’ be arranged
(a) if the two O’s must not come together?
(b) if the consonants and vowels must occupy alternate places?

Answer 1

Hint: The question revolves around the concepts of permutations and combinations. The question has two subparts and each subpart is to be solved in a different way. According to the fundamental theorem of counting, if there are p ways of doing one thing and q ways of doing another thing, then there are $p \times q$ ways of doing both the things. So, we find out the number of options in each condition and then multiply the number of options so as to get the required answer.

Complete step-by-step solution:
So, we are required to find the number of ways in which the letters of the word ‘SALOON’ can be arranged so that the two O’s must not come together.
So, in order to find the number of arrangements if the word ‘SALOON’ with two O’s not coming together, we first find the total number of arrangements of the given word and then subtract the arrangements in which the two O’s come together so as to find the final answer to the question.
The number of letters in the word ‘SALOON’ is $6$.
Also, the letter O is repeated twice in the word.
So, the number of total arrangements of the letters of the word ‘SALOON’ is $\dfrac{{6!}}{{2!}}$.
Now, we calculate the arrangements in which the two O’s come together.
So, we form a separate bag consisting of two O’s and consider it to be one single entity. Then, we find the arrangements of the letters.
So, the number of entities to be arranged including the bag consisting of two O’s is $4 + 1 = 5$.
Now, the number of ways in which these five entities can be arranged is $5!$. Since there are identical letters in the bag consisting of O’s. So, the arrangement of letters in the bag is of no significance.
So, the total number of arrangements in which the two O’s come together is $5!$.
Therefore, the number of ways in which the letters of the word ‘SALOON’ can be arranged so that the two O’s must not come together is $\dfrac{{6!}}{{2!}} - 5!$
Substituting in the values of factorials, we get,
$ \Rightarrow \dfrac{{720}}{2} - 120$
Cancelling the common factors in numerator and denominator and simplifying the calculations, we get,
$ \Rightarrow 360 - 120 = 240$
So, the number of ways in which the letters of the word ‘SALOON’ can be arranged so that the two O’s must not come together is $240$.
Now, we have to find the number of ways the letters of word ‘SALOON’ can be arranged so that the consonants and vowels must occupy alternate spaces.
The number of vowels in the word ‘SALOON’ is $3$.
The number of consonants in the word ‘SALOON’ is $3$.
Now, we know that all the consonants in the word ‘SALOON’ are different. But the vowel O is repeated twice in the word ‘SALOON’.
Now, the number of ways of arranging vowels of the word ‘SALOON’ is $\dfrac{{3!}}{{2!}}$.
Also, the number of ways of arranging consonants of the word ‘SALOON’ is $3!$.
Now, we know that either the vowels occupy the odd places and consonants occupy the even places or vice versa. So, there are a total of $2!$ ways of choosing the places for consonants and vowels.
Hence, the number of ways of arranging the letters of the word ‘SALOON’ so that the consonants and vowels must occupy alternate spaces is $\dfrac{{3!}}{{2!}} \times 3! \times 2!$
Substituting the values of factorials and doing the calculations, we get,
$ \Rightarrow \dfrac{6}{2} \times 6 \times 2$
$ \Rightarrow 36$
So, the number of ways of arranging the letters of the word ‘SALOON’ so that the consonants and vowels must occupy alternate spaces is $36$.

Note: Combination means choosing elements is only that matters, whereas permutation is an ordered combination. The formula used to find combination is \[{}^n{C_r} = \dfrac{{n!}}{{r! \times \left( {n - r} \right)!}}\] . Permutation is a method used to calculate the total outcome of a situation where order is important, the formula used to find permutation is and \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\].