If number of arrangements of the letters of the word MONOTONICITY in which O's do not appear adjacentely is 9! (k), then k equals

Question

If number of arrangements of the letters of the word MONOTONICITY in which O's do not appear adjacentely is 9! (k), then k equals (A) 12 (B) 15 (C) 10 (D) 8

Tardigrade Admin · Accepted Answer

O = 3, M = 1, C = 1, Y = 1, N = 2, T = 2, I = 2 (9 !/2 ! ⋅ 2 ! ⋅ 2 !) × 10 C 3=15 × 9 ! ⇒ k =15

Q. If number of arrangements of the letters of the word MONOTONICITY in which O's do not appear adjacentely is $9! (k)$ , then k equals

12

15

10

8

$O = 3, M = 1, C = 1, Y = 1, N = 2, T = 2, I = 2$
$\frac{9 !}{2 ! \cdot 2 ! \cdot 2 !} \times^{10} C_{3} = 15 \times 9!$
$\Rightarrow k = 15$

Q. If number of arrangements of the letters of the word MONOTONICITY in which O's do not appear adjacentely is 9!(k), then k equals

12

15

10

8

O=3,M=1,C=1,Y=1,N=2,T=2,I=2 2!⋅2!⋅2!9!​×10C3​=15×9! ⇒k=15

Q. If number of arrangements of the letters of the word MONOTONICITY in which O's do not appear adjacentely is $9! (k)$ , then k equals

$O = 3, M = 1, C = 1, Y = 1, N = 2, T = 2, I = 2$
$\frac{9 !}{2 ! \cdot 2 ! \cdot 2 !} \times^{10} C_{3} = 15 \times 9!$
$\Rightarrow k = 15$