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How many $4$ letter words with or without meaning can be formed using the letters of the word RUCHIKA, if
$a)$ Repetition of letters is not allowed.
$b)$ Repetition of letters is allowed.

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Last updated date: 12th Jul 2024
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Answer
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Hint: First we have to define what the terms we need to solve the problem are.
Since we need to find the framing of the word RUCHIKA into the four-letter words may or may not be meaningful but with or without repetition of the same letters.
Like repetition not allowed means once the letter U in the word RUCHIKA is arrived then we cannot use it again to frame another set of words.


Complete step-by-step solution:
Since there are total seven letters in the words RUCHIKA, so $7$ is the total value and we need only four letters words that means $4$ is the restriction and thus ${}^7{c_4}$ is the combination that number of ways to put the set of seven letter into the four letters words.
$a)$Repetition of letters is not allowed.
Since the repetition of the letter is not allowed so once a letter arrived in a word will need to be eliminated for the other words and thus, we already know the combination is ${}^7{c_4}$ is the overall combination and four letters can be arrived at word one thus we have four ways;
And similarly for the word two we have (three letter as without repetition)
For the word three we have two letters and finally for the last word we have only one letter left
Thus, we get the required solutions as ${}^7{c_4} \times 4 \times 3 \times 2 \times 1$(applying the multiplication formula)
Thus, we get ${}^7{c_4} \times 4 \times 3 \times 2 \times 1 = 840$ is the repetition not allowed letters
$b)$ Repetition of letters is allowed.
If the reputation of the letters is allowed then it will be very simple; for the first word we have seven letters thus $7$ ways, and for the second word we also have the same $7$ letters as repetition is now allowed as similarly for the third and final words, we have the $7$ letters then of the combination
Thus, we get $7 \times 7 \times 7 \times 7$(repetition is allowed)
Therefore, we get $7 \times 7 \times 7 \times 7 = 2401$ of repetition of the letters for the four words.

Note: Combination is the number of ways to arrange or count the given problem and ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$
n is the total amount and r is the object that needs to arrange into the different sets. And repetition means that the same letter can be able to come again and again in all the sets.