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,
A) Contain the letter S.
B) Do not contain the letter O.
C) Consist of a vowel between two consonants.

Answer 1

Hint: Since each letter of the word “DOUBLES” are single. So, all the three-letter code words will be different. In the first part, the letter S must be fixed in any place or position of the three-letter code words. In the second part, we must exclude the letter O and find the three-letter code words. In the last part, we will fix consonants at the first and third positions and vowels in the second position.

Complete step-by-step answer:
Given:- Seven cards, each bearing a different letter.
For the total number of three letters code-words. Seven possible cards can be placed for the first position. After the first position, six cards are left. So, 6 possible cards can be placed for the second position. Similarly, 5 possible cards can be placed for the third position.
Then, the total three-letter code words that can be formed is,
$7 \times 6 \times 5 = 210$
Now, we have to find a three-letter code word that contains the letter S.
There are 3 possible positions for the card S. As one position is fixed for a card with letter S, the two-position are left with 6 cards. So, 6 possible cards can be placed for the second position. Similarly, 5 possible cards can be placed for the third position.
Then, the total three-letter code words that can be formed is,
$3 \times 6 \times 5 = 90$
Now, we have to find a three-letter code word which does not contain the letter O.
Since the card with the letter O will not be used. So, six possible cards can be placed for the first position. After the first position, five cards are left. So, five possible cards can be placed for the second position. Similarly, four possible cards can be placed for the third position.
Then, the total three-letter code words that can be formed is,
$6 \times 5 \times 4 = 120$
Now, we have to find a three-letter code word which consists of a vowel between two consonants.
Since there are 3 vowels and 4 consonants. So, four possible cards can be placed for the first position. After the first position, three consonant cards are left. So, three possible cards can be placed for the third position. There are three vowel cards. So, three possible cards can be placed for the second position.
Then, the total three-letter code words that can be formed is,
$4 \times 3 \times 3 = 36$
Hence, the total number of three letters code-words are $210$, the three-letter code word which contains the letter S is $90$, the three-letter code word which does not contain the letter O are $120$ and the three-letter code word which consists of a vowel between two consonants are $36$.

Note: The students must remember that every card has different letters and 1 card will be used at a time for 1 position. So, he/she must not use the repetition of letters to solve this type of problem.
This type of problem can be solved using a combination formula. For three-letter words used \[{}^7{C_3}\] . But the three letters must be arranged at different positions, so use $3!$. So, the final answer will be calculated by \[{}^7{C_3} \times 3!\].