The number of ways in which alphabets of the word MATHOPHOBIA can be arranged so that no two identical vowels are separated and no two identical consonants occur together is

Question

The number of ways in which alphabets of the word MATHOPHOBIA can be arranged so that no two identical vowels are separated and no two identical consonants occur together is (A) (8 !/2) (B) 8 ! (C) (7 !/2) (D) (7 × 8 !/2)

Tardigrade Admin · Accepted Answer

× M × AA × OO × P × B × I × T × 7 ! × 8 C 2=(8 ! × 7/2)

Q. The number of ways in which alphabets of the word MATHOPHOBIA can be arranged so that no two identical vowels are separated and no two identical consonants occur together is

$\frac{8 !}{2}$

8 !

$\frac{7 !}{2}$

$\frac{7 \times 8 !}{2}$

$\times M \times AA \times OO \times P \times B \times I \times T \times$
$7! \times^{8} C_{2} = \frac{8 ! \times 7}{2}$

Q. The number of ways in which alphabets of the word MATHOPHOBIA can be arranged so that no two identical vowels are separated and no two identical consonants occur together is

28!​

8 !

27!​

27×8!​

×M×AA×OO×P×B×I×T× 7!×8C2​=28!×7​

$\frac{8 !}{2}$

$\frac{7 !}{2}$

$\frac{7 \times 8 !}{2}$

$\times M \times AA \times OO \times P \times B \times I \times T \times$
$7! \times^{8} C_{2} = \frac{8 ! \times 7}{2}$