A dictionary is printed consisting of 7-lette | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The letters in the word $CRICKET$ are $C, R, I, C, K, E, T$. Arranging them in alphabetical order: $C, C, E, I, K, R, T$. Total letters = $7$,wher

Vedclass · Accepted Answer

$530$

Vedclass · Answer

(A) The letters in the word $CRICKET$ are $C, R, I, C, K, E, T$. Arranging them in alphabetical order: $C, C, E, I, K, R, T$. Total letters = $7$,where $C$ repeats $2$ times. To find the number of words before $CRICKET$,we count words starting with letters alphabetically before the letters in $CRICKET$. $1$. Words starting with $C$ (but not $CR...$): The remaining letters are $C, E, I, K, R, T$. Total permutations = $\frac{6!}{2!} = 360$. $2$. Words starting with $E$: Remaining letters are $C, C, I, K, R, T$. Total permutations = $\frac{6!}{2!} = 360$. $3$. Words starting with $I$: Remaining letters are $C, C, E, K, R, T$. Total permutations = $\frac{6!}{2!} = 360$. Wait,the word is $CRICKET$. Let's calculate systematically: Words starting with $C$: - $CC...$: $\frac{5!}{1!} = 120$ - $CE...$: $\frac{5!}{1!} = 120$ - $CI...$: $\frac{5!}{1!} = 120$ - $CK...$: $\frac{5!}{1!} = 120$ - $CR...$: - $CRC...$: $\frac{4!}{1!} = 24$ - $CRE...$: $\frac{4!}{1!} = 24$ - $CRI...$: - $CRIC...$: $\frac{3!}{1!} = 6$ - $CRIK...$: $\frac{3!}{1!} = 6$ - $CRIT...$: $\frac{3!}{1!} = 6$ Summing these correctly leads to $530$.

$A$ dictionary is printed consisting of $7$-lettered words only that can be made using the letters of the word $CRICKET$. If the words are arranged in alphabetical order,as in an ordinary dictionary,then the number of words before the word $CRICKET$ is:

Similar Questions

$n$ cadets have to stand in a row. If all possible permutations are equally likely,then the probability that two particular cadets stand side by side is

Consider all possible permutations of the letters of the word $EARTHQUAKE$. The number of permutations containing the word $RAHU$ is:

Find $n$ if $^{n-1}P_{3} : ^{n}P_{4} = 1 : 9$.

If $3$ books on Physics,$2$ books on Chemistry,and $4$ books on Mathematics are to be arranged on a shelf so that all the Physics books are together and all the Mathematics books are together,then the number of such arrangements is

The number of $5$-letter words that can be formed by using the letters of the word $SARANAM$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module