Answer: Option B. -> 144
THE WORD MEADOWS HAS 7 LETTERS OF WHICH 3 ARE VOWELS.
-V-V-V-
AS THE VOWELS HAVE TO OCCUPY EVEN PLACES, THEY CAN BE ARRANGED IN THE 3 EVEN PLACES IN 3! I.E., 6 WAYS. WHILE THE CONSONANTS CAN BE ARRANGED AMONG THEMSELVES IN THE REMAINING 4 PLACES IN 4! I.E., 24 WAYS.
HENCE THE TOTAL WAYS ARE 24 * 6 = 144.

Answer: Option B. -> 8!
TOTAL NUMBER OF LETTERS = 8
USING THESE LETTERS THE NUMBER OF 8 LETTERS WORDS FORMED IS ⁸P₈ = 8!.

Answer: Option D. -> 108
THE BOY CAN SELECT ONE TROUSER IN NINE WAYS.
THE BOY CAN SELECT ONE SHIRT IN 12 WAYS.
THE NUMBER OF WAYS IN WHICH HE CAN SELECT ONE TROUSER AND ONE SHIRT IS 9 * 12 = 108 WAYS.

Answer: Option B. -> 435
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!

Answer: Option C. -> 120960
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!

Answer: Option B. -> 720
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!

Answer: Option B. -> 6
THERE ARE FIVE LETTERS IN THE GIVEN WORD.
CONSIDER 5 BLANKS ….
THE FIRST BLANK AND LAST BLANK MUST BE FILLED WITH N AND A ALL THE REMAINING THREE BLANKS CAN BE FILLED WITH THE REMAINING 3 LETTERS IN 3! WAYS.
THE NUMBER OF WORDS = 3! = 6.

Answer: Option D. -> 24
THE NUMBER OF LETTERS IN THE GIVEN WORD IS FOUR.
THE NUMBER OF THREE LETTER WORDS THAT CAN BE FORMED USING THESE FOUR LETTERS IS ⁴P₃ = 4 * 3 * 2 = 24.

Answer: Option C. -> 24
CASE I : 2 BALLS OF THE SAME COLOUR AND TWO BALLS ARE A DIFFERENT COLOUR ARE ARRANGED.
TWO BALLS OF THE SAME COLOUR AND TWO BALLS OF DIFFERENT COLOURS CAN BE ARRANGED TOGETHER IN WHICH TWO BALLS OF THE SAME COLOUR ARE ADJACENT =4!/2!X2! = 6 WAYS
THEREFORE, TOTAL NUMBER OF ARRANGEMENTS = 6×3 =18 WAYS
CASE II : TWO COLOURS OUT OF 3 CAN BE SELECTED IN = 3C1 = 3WAYS
NOW 2 BALLS OF EACH COLOUR CAN BE ARRANGED ALTERNATIVELY IN 2 WAYS
THUS 4 BALLS CAN BE ARRANGED(TWO OF EACH COLOURS)
= 3×2 = 6WAYS
HENCE TOTAL NUMBER OF ARRANGEMENTS = 18+6 =24 WAYS

Answer: Option A. -> 63
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!