The number of words,with or without meaning,t | vedclass

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

Question

(C) The word $ASSASSINATION$ contains $13$ letters: $A, S, S, A, S, S, I, N, A, T, I, O, N$.Vowels are: $A, A, A, I, I, O$ (Total $6$ vowels).Consonan

Vedclass · Accepted Answer

$50400$

Vedclass · Answer

(C) The word $ASSASSINATION$ contains $13$ letters: $A, S, S, A, S, S, I, N, A, T, I, O, N$.Vowels are: $A, A, A, I, I, O$ (Total $6$ vowels).Consonants are: $S, S, S, S, N, N, T$ (Total $7$ consonants).Treating the $6$ vowels as a single unit,we have $7$ consonants + $1$ unit = $8$ items to arrange.Number of arrangements of these $8$ items (where $S$ repeats $4$ times and $N$ repeats $2$ times) is $\frac{8!}{4!2!} = \frac{40320}{24 	imes 2} = 840$.Now,arrange the $6$ vowels within the unit (where $A$ repeats $3$ times and $I$ repeats $2$ times):Number of arrangements = $\frac{6!}{3!2!} = \frac{720}{6 	imes 2} = 60$.Total number of words = $840 	imes 60 = 50400$.

The number of words,with or without meaning,that can be formed using all the letters of the word $ASSASSINATION$ so that the vowels occur together,is $.............$.

Similar Questions

$A$ three-digit number $n$ is such that the last two digits are equal and differ from the first digit. The number of such $n$'s is:

If $7 \cdot ^nP_3 = 20 \cdot ^{n+1}P_2$,then $n = \dots$

$_{n-1}P_r + r \cdot (_{n-1}P_{r-1}) = \dots$

How many five-digit numbers can be formed using the digits $2, 0, 4, 3, 8$ if repetition of digits is not allowed?

Six objects $O_1$ to $O_6$ are arranged one on top of the other. In how many ways can these be arranged such that $O_1$ and $O_2$ are the $2$ bottom-most objects?

Vedclass Test Series

Exam Paper Generator

Online Exam Module