Quantitative Aptitude
PERMUTATION AND COMBINATION MCQs
Permutations And Combinations
Total Questions : 444
| Page 41 of 45 pages
Answer: Option D. -> 30
THE WORD CONTAINS FIVE CONSONANTS. THREE VOWELS, THREE CONSONANTS CAN BE SELECTED FROM FIVE CONSONANTS IN ⁵C₃ WAYS, TWO VOWELS CAN BE SELECTED FROM THREE VOWELS IN ³C₂ WAYS.
3 CONSONANTS AND 2 VOWELS CAN BE SELECTED IN ⁵C₂ . ³C₂ WAYS I.E., 10 * 3 = 30 WAYS.
THE WORD CONTAINS FIVE CONSONANTS. THREE VOWELS, THREE CONSONANTS CAN BE SELECTED FROM FIVE CONSONANTS IN ⁵C₃ WAYS, TWO VOWELS CAN BE SELECTED FROM THREE VOWELS IN ³C₂ WAYS.
3 CONSONANTS AND 2 VOWELS CAN BE SELECTED IN ⁵C₂ . ³C₂ WAYS I.E., 10 * 3 = 30 WAYS.
Answer: Option A. -> 60
THE GIVEN DIGITS ARE 1, 2, 3, 5, 7, 9
A NUMBER IS EVEN WHEN ITS UNITS DIGIT IS EVEN. OF THE GIVEN DIGITS, TWO IS THE ONLY EVEN DIGIT.
UNITS PLACE IS FILLED WITH ONLY ‘2’ AND THE REMAINING THREE PLACES CAN BE FILLED IN ⁵P₃ WAYS.
NUMBER OF EVEN NUMBERS = ⁵P₃ = 60.
THE GIVEN DIGITS ARE 1, 2, 3, 5, 7, 9
A NUMBER IS EVEN WHEN ITS UNITS DIGIT IS EVEN. OF THE GIVEN DIGITS, TWO IS THE ONLY EVEN DIGIT.
UNITS PLACE IS FILLED WITH ONLY ‘2’ AND THE REMAINING THREE PLACES CAN BE FILLED IN ⁵P₃ WAYS.
NUMBER OF EVEN NUMBERS = ⁵P₃ = 60.
Answer: Option A. -> 360
THE GIVEN DIGITS ARE SIX.
THE NUMBER OF FOUR DIGIT NUMBERS THAT CAN BE FORMED USING SIX DIGITS IS ⁶P₄ = 6 * 5 * 4 * 3 = 360.
THE GIVEN DIGITS ARE SIX.
THE NUMBER OF FOUR DIGIT NUMBERS THAT CAN BE FORMED USING SIX DIGITS IS ⁶P₄ = 6 * 5 * 4 * 3 = 360.
Answer: Option C. -> ¹⁶C₂
TOTAL NUMBER OF BALLS = 9 + 3 + 4
TWO BALLS CAN BE DRAWN FROM 16 BALLS IN ¹⁶C₂ WAYS.
TOTAL NUMBER OF BALLS = 9 + 3 + 4
TWO BALLS CAN BE DRAWN FROM 16 BALLS IN ¹⁶C₂ WAYS.
Answer: Option B. -> 500
WE CAN SELECT ONE BOY FROM 20 BOYS IN 20 WAYS.
WE SELECT ONE GIRL FROM 25 GIRLS IN 25 WAYS
WE SELECT A BOY AND GIRL IN 20 * 25 WAYS I.E., = 500 WAYS.
WE CAN SELECT ONE BOY FROM 20 BOYS IN 20 WAYS.
WE SELECT ONE GIRL FROM 25 GIRLS IN 25 WAYS
WE SELECT A BOY AND GIRL IN 20 * 25 WAYS I.E., = 500 WAYS.
Answer: Option A. -> 165
TOTAL NUMBER OF PERSONS IN THE COMMITTEE = 5 + 6 = 11
NUMBER OF WAYS OF SELECTING GROUP OF EIGHT PERSONS = ¹¹C₈ = ¹¹C₃ = (11 * 10 * 9)/(3 * 2) = 165 WAYS.
TOTAL NUMBER OF PERSONS IN THE COMMITTEE = 5 + 6 = 11
NUMBER OF WAYS OF SELECTING GROUP OF EIGHT PERSONS = ¹¹C₈ = ¹¹C₃ = (11 * 10 * 9)/(3 * 2) = 165 WAYS.
Answer: Option C. -> 215
SINCE EACH RING CONSISTS OF SIX DIFFERENT LETTERS, THE TOTAL NUMBER OF ATTEMPTS POSSIBLE WITH THE THREE RINGS IS = 6 * 6 * 6 = 216. OF THESE ATTEMPTS, ONE OF THEM IS A SUCCESSFUL ATTEMPT.
MAXIMUM NUMBER OF UNSUCCESSFUL ATTEMPTS = 216 – 1 = 215.
SINCE EACH RING CONSISTS OF SIX DIFFERENT LETTERS, THE TOTAL NUMBER OF ATTEMPTS POSSIBLE WITH THE THREE RINGS IS = 6 * 6 * 6 = 216. OF THESE ATTEMPTS, ONE OF THEM IS A SUCCESSFUL ATTEMPT.
MAXIMUM NUMBER OF UNSUCCESSFUL ATTEMPTS = 216 – 1 = 215.
Answer: Option C. -> 30
THE NUMBER OF WAYS TO SELECT THREE MEN AND TWO WOMEN SUCH THAT ONE MAN AND ONE WOMAN ARE ALWAYS SELECTED = NUMBER OF WAYS SELECTING TWO MEN AND ONE WOMAN FROM MEN AND FIVE WOMEN
= ⁴C₂ * ⁵C₁ = (4 * 3)/(2 * 1) * 5
= 30 WAYS.
THE NUMBER OF WAYS TO SELECT THREE MEN AND TWO WOMEN SUCH THAT ONE MAN AND ONE WOMAN ARE ALWAYS SELECTED = NUMBER OF WAYS SELECTING TWO MEN AND ONE WOMAN FROM MEN AND FIVE WOMEN
= ⁴C₂ * ⁵C₁ = (4 * 3)/(2 * 1) * 5
= 30 WAYS.
Answer: Option B. -> 200
THE NUMBER OF WAYS TO SELECT TWO MEN AND THREE WOMEN = ⁵C₂ * ⁶C₃
= (5 *4 )/(2 * 1) * (6 * 5 * 4)/(3 * 2)
= 200
THE NUMBER OF WAYS TO SELECT TWO MEN AND THREE WOMEN = ⁵C₂ * ⁶C₃
= (5 *4 )/(2 * 1) * (6 * 5 * 4)/(3 * 2)
= 200
Answer: Option C. -> 5040
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
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