Quantitative Aptitude
PERMUTATION AND COMBINATION MCQs
Permutations And Combinations
Total Questions : 444
| Page 45 of 45 pages
Answer: Option C. -> 119
NUMBER OF WORDS WHICH CAN BE FORMED = 5! – 1 = 120 – 1 = 119.
NUMBER OF WORDS WHICH CAN BE FORMED = 5! – 1 = 120 – 1 = 119.
Answer: Option A. -> 20
THERE ARE THREE LADIES AND FIVE GENTLEMEN AND A COMMITTEE OF 5 MEMBERS TO BE FORMED.
NUMBER OF WAYS SUCH THAT TWO LADIES ARE ALWAYS INCLUDED IN THE COMMITTEE = ⁶C₃ = (6 * 5 * 4)/6 = 20.
THERE ARE THREE LADIES AND FIVE GENTLEMEN AND A COMMITTEE OF 5 MEMBERS TO BE FORMED.
NUMBER OF WAYS SUCH THAT TWO LADIES ARE ALWAYS INCLUDED IN THE COMMITTEE = ⁶C₃ = (6 * 5 * 4)/6 = 20.
Answer: Option C. -> 4320
X NOT YOUNGER_______ ↑
THE LAST BALL CAN BE THROWN BY ANY OF THE REMAINING 6 PLAYERS. THE FIRST 6 PLAYERS CAN THROW THE BALL IN ⁶P₆ WAYS.
THE REQUIRED NUMBER OF WAYS = 6(6!) = 4320
X NOT YOUNGER_______ ↑
THE LAST BALL CAN BE THROWN BY ANY OF THE REMAINING 6 PLAYERS. THE FIRST 6 PLAYERS CAN THROW THE BALL IN ⁶P₆ WAYS.
THE REQUIRED NUMBER OF WAYS = 6(6!) = 4320
Answer: Option C. -> 2880
TREAT ALL BOYS AS ONE UNIT. NOW THERE ARE FOUR STUDENTS AND THEY CAN BE ARRANGED IN 4! WAYS. AGAIN FIVE BOYS CAN BE ARRANGED AMONG THEMSELVES IN 5! WAYS.
REQUIRED NUMBER OF ARRANGEMENTS = 4! * 5! = 24 * 120 = 2880.
TREAT ALL BOYS AS ONE UNIT. NOW THERE ARE FOUR STUDENTS AND THEY CAN BE ARRANGED IN 4! WAYS. AGAIN FIVE BOYS CAN BE ARRANGED AMONG THEMSELVES IN 5! WAYS.
REQUIRED NUMBER OF ARRANGEMENTS = 4! * 5! = 24 * 120 = 2880.
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