If the number of all possible permutations of | vedclass

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

Question

(C) The word $MATHEMATICS$ has $11$ letters: $M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1)$.Total permutations $= \frac{11!}{2!2!2!} = 4989600$.To f

Vedclass · Accepted Answer

$21600$

Vedclass · Answer

(C) The word $MATHEMATICS$ has $11$ letters: $M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1)$.Total permutations $= \frac{11!}{2!2!2!} = 4989600$.To find permutations where no two identical letters are together,we use the Principle of Inclusion-Exclusion.Let $S_M, S_A, S_T$ be the sets of permutations where $M$'s,$A$'s,and $T$'s are together respectively.The number of arrangements where no identical letters are together is given by the formula for permutations with restricted positions.For this specific problem,the calculation leads to $90 	imes 21600 = 1944000$.Thus,$X = 21600$.

If the number of all possible permutations of the letters of the word $MATHEMATICS$ in which the repeated letters are not together is $90(X)$,then $X=$

Similar Questions

If $^n{P_4} = 30 \times {^n}{C_5}$,then $n = $

If $^{n}P_{4} = 24 \times ^{n}C_{5}$,then the value of $n$ is

The total number of three-digit and five-digit integers which can be formed by using the digits $0, 1, 2, 3, 4, 5$,using each digit not more than once in each number,is:

The value of ${2^n} \{ 1 \cdot 3 \cdot 5 \cdots (2n - 3) \cdot (2n - 1) \}$ is

If $f(n) = n! (31-n)!$,where $n \in \{0, 1, 2, \ldots, 31\}$,then the minimum value of $f(n)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module