Question

Seven cards, each bearing a letter, can be arranged to spell the word “DOUBLES”. How many three letters code-words can be formed from these cards? How many of these words:
(i) contain the letter S (ii) do not contain the letter O (iii) consist of a vowel between two consonants ?

Answer 1

- Each letter of the word “DOUBLES” has a single card. So, all the letters of the three-letter code word will be different. So, in each case we had to fix the given word (like S) at any position of the 3-letters word and then find the number of possible letters for all the other positions.

Complete step by step answer :
Let us first find the three letters code word that can be formed without any condition.
As we know that there are seven cards (i.e. D, O, U, B, L, E, S) out of which we had to choose any three cards to make a three letters word.
So, as we know that three are three positions in the three letters words.
So, let us place cards on each position. So, for first position three are seven possible cards that can be placed. Now for the second position there will be only six possible cards left because each card can be used at one position. Now for the last position there will be a total of five cards left because out of seven two cards were already placed at first and second position.
So, total three letters code word that can be formed will be equal to 765 = 210.
(i) Now we had to find a three letter code-word which contains ‘S’ as one of its letters.
So, first we had to fix the card with letter S and then find the number of possible cards for the other two positions.
Now there are three possible positions for card ‘S’.
Now we are left with two positions (one is filled with S) and 6 (i.e. D, O, U, B, L and E) cards.
So, there are 6 possible cards for the second position.
And for the third position there will be 5 cards left (‘S’ and one other card is used).
Hence, the total three-letters words that can be formed having one card of letter S will be equal to 365 = 90.
(ii) Now we have to find a three letter code-word which does not contain ‘O’ as one of its letters.
So, there are only six cards (D, U, B, L, E, S) left that can be chosen for making three letters.
So, let us place cards on each position. So, for first position three are six possible cards that can be placed. Now for the second position there will be only five possible cards left because each card can be used at one position. Now for the last position there will be a total of four cards left because out of seven two cards were already placed at first and second position and one card is for letter O.
Hence, the total three-letters words that can be formed that does not contain card with letter O will be equal to 654 = 120.
(iii) To find the number of possible words that consist of a vowel between two consonants. First, we had to find the number of cards with consonants and vowels.
As we know that in English alphabet only A, E, I, O, U are the vowels. All other alphabets are consonants.
So, there will be three cards (O, U, E) of vowels and four cards (D, B, L, S) of consonants.
So, now there are 4 possible cards (consonants) available for first position. Then 3 possible cards (vowels) for second position. And after that three possible cards for the last position (consonants).
Hence, the total three-letters words that can be formed having vowel between two consonants will be equal to 433 = 36

Note: Whenever we come up with this type of problem then there is another way to find the number of possible three letter words and that can be done by using a combination formula, like for three letter words we had to choose three cards from 7 available cards. So, we can use ${}^7{C_3}$ but here we had to also arrange all the letters. So, the final answer will be ${}^7{C_3} \times 3!$ because to arrange n letters we multiply by $n!$.