Quantitative Aptitude
PERMUTATION AND COMBINATION MCQs
Permutations And Combinations
Total Questions : 444
| Page 44 of 45 pages
Answer: Option C. -> 624
MULTIPLE CHOICE TYPE QUESTIONS = 1 2 3 4
TOTAL NUMBER OF WAYS = 5X5X5X5 =625.
A NUMBER OF CORRECT ANSWER = 1.
NUMBER OF FALSE ANSWERS = 625-1 =624.
MULTIPLE CHOICE TYPE QUESTIONS = 1 2 3 4
TOTAL NUMBER OF WAYS = 5X5X5X5 =625.
A NUMBER OF CORRECT ANSWER = 1.
NUMBER OF FALSE ANSWERS = 625-1 =624.
Answer: Option D. -> 480
1. AS THERE ARE SIX PLAYERS, SO TOTAL WAYS IN WHICH THEY CAN BE ARRANGED = 6!WAYS =720.
A NUMBER OF WAYS IN WHICH ASIM AND RAHEEM ARE TOGETHER = 5!X2 = 240.
THEREFORE, NUMBER OF WAYS WHEN THEY DON’T REMAIN TOGETHER = 720 -240 =480.
1. AS THERE ARE SIX PLAYERS, SO TOTAL WAYS IN WHICH THEY CAN BE ARRANGED = 6!WAYS =720.
A NUMBER OF WAYS IN WHICH ASIM AND RAHEEM ARE TOGETHER = 5!X2 = 240.
THEREFORE, NUMBER OF WAYS WHEN THEY DON’T REMAIN TOGETHER = 720 -240 =480.
Answer: Option D. -> 1024
EACH QUESTION CAN BE ANSWERED IN 2 WAYS.
10 QUESTIONS CAN BE ANSWERED = 2 POWER OF 10= 1024 WAYS.
EACH QUESTION CAN BE ANSWERED IN 2 WAYS.
10 QUESTIONS CAN BE ANSWERED = 2 POWER OF 10= 1024 WAYS.
Answer: Option C. -> 7
MAXIMUM NUMBER OF SUCH DIFFERENT GROUPS = ABC, ABD,ABE, BCE,BDE,CEA,DEA =7.
ALTERNATE METHOD:
TOTAL NUMBER OF WAY IN WHICH 3 BOYS CAN BE SELECTED OUT OF 5 IS 5C3
NUMBER OF WAYS IN WHICH CD COMES TOGETHER = 3 (CDA,CDB,CDE)
THEREFORE, REQUIRED NUMBER OF WAYS = 5C3 -3
= 10-3 =7.
MAXIMUM NUMBER OF SUCH DIFFERENT GROUPS = ABC, ABD,ABE, BCE,BDE,CEA,DEA =7.
ALTERNATE METHOD:
TOTAL NUMBER OF WAY IN WHICH 3 BOYS CAN BE SELECTED OUT OF 5 IS 5C3
NUMBER OF WAYS IN WHICH CD COMES TOGETHER = 3 (CDA,CDB,CDE)
THEREFORE, REQUIRED NUMBER OF WAYS = 5C3 -3
= 10-3 =7.
Answer: Option D. -> ²²C₁₀ – 1
THE TOTAL NUMBER OF WAYS OF FORMING THE GROUP OF TEN REPRESENTATIVES IS ²²C₁₀.
THE TOTAL NUMBER OF WAYS OF FORMING THE GROUP THAT CONSISTS OF NO SENIORS IS ¹⁰C₁₀ = 1 WAY
THE REQUIRED NUMBER OF WAYS = ²²C₁₀ – 1
THE TOTAL NUMBER OF WAYS OF FORMING THE GROUP OF TEN REPRESENTATIVES IS ²²C₁₀.
THE TOTAL NUMBER OF WAYS OF FORMING THE GROUP THAT CONSISTS OF NO SENIORS IS ¹⁰C₁₀ = 1 WAY
THE REQUIRED NUMBER OF WAYS = ²²C₁₀ – 1
Answer: Option B. -> 32
WE KNOW THAT, THE NUMBER OF STRAIGHT LINES THAT CAN BE FORMED BY THE 11 POINTS IN WHICH 6 POINTS ARE COLLINEAR AND NO OTHER SET OF THREE POINTS, EXCEPT THOSE THAT CAN BE SELECTED OUT OF THESE 6 POINTS ARE COLLINEAR.
HENCE, THE REQUIRED NUMBER OF STRAIGHT LINES
= ¹¹C₂ – ⁶C₂ – ⁵C₂ + 1 + 1
= 55 – 15 – 10 + 2 = 32
WE KNOW THAT, THE NUMBER OF STRAIGHT LINES THAT CAN BE FORMED BY THE 11 POINTS IN WHICH 6 POINTS ARE COLLINEAR AND NO OTHER SET OF THREE POINTS, EXCEPT THOSE THAT CAN BE SELECTED OUT OF THESE 6 POINTS ARE COLLINEAR.
HENCE, THE REQUIRED NUMBER OF STRAIGHT LINES
= ¹¹C₂ – ⁶C₂ – ⁵C₂ + 1 + 1
= 55 – 15 – 10 + 2 = 32
Answer: Option C. -> ¹²C₇ * ¹⁰C₅
HERE, FIVE SENIORS OUT OF 12 SENIORS CAN BE SELECTED IN ¹²C₅ WAYS. ALSO, FIVE JUNIORS OUT OF TEN JUNIORS CAN BE SELECTED ¹⁰C₅ WAYS. HENCE THE TOTAL NUMBER OF DIFFERENT WAYS OF SELECTION = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
= ¹²C₅ = ¹²C₇
HERE, FIVE SENIORS OUT OF 12 SENIORS CAN BE SELECTED IN ¹²C₅ WAYS. ALSO, FIVE JUNIORS OUT OF TEN JUNIORS CAN BE SELECTED ¹⁰C₅ WAYS. HENCE THE TOTAL NUMBER OF DIFFERENT WAYS OF SELECTION = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
= ¹²C₅ = ¹²C₇
Answer: Option C. -> 600
THE NUMBER OF WAYS OF SELECTING THREE MEN, TWO WOMEN AND THREE CHILDREN IS:
= ⁴C₃ * ⁶C₂ * ⁵C₃
= (4 * 3 * 2)/(3 * 2 * 1) * (6 * 5)/(2 * 1) * (5 * 4 * 3)/(3 * 2 * 1)
= 4 * 15 * 10
= 600 WAYS.
THE NUMBER OF WAYS OF SELECTING THREE MEN, TWO WOMEN AND THREE CHILDREN IS:
= ⁴C₃ * ⁶C₂ * ⁵C₃
= (4 * 3 * 2)/(3 * 2 * 1) * (6 * 5)/(2 * 1) * (5 * 4 * 3)/(3 * 2 * 1)
= 4 * 15 * 10
= 600 WAYS.
Answer: Option B. -> 1440
IN THE WORD, “MATERIAL” THERE ARE THREE VOWELS A, I, E.
IF ALL THE VOWELS ARE TOGETHER, THE ARRANGEMENT IS MTRL’AAEI’.
CONSIDER AAEI AS ONE UNIT. THE ARRANGEMENT IS AS FOLLOWS.
M T R L A A E I
THE ABOVE 5 ITEMS CAN BE ARRANGED IN 5! WAYS AND AAEI CAN BE ARRANGED AMONG THEMSELVES IN 4!/2! WAYS.
NUMBER OF REQUIRED WAYS OF ARRANGING THE ABOVE LETTERS = 5! * 4!/2!
= (120 * 24)/2 = 1440 WAYS.
IN THE WORD, “MATERIAL” THERE ARE THREE VOWELS A, I, E.
IF ALL THE VOWELS ARE TOGETHER, THE ARRANGEMENT IS MTRL’AAEI’.
CONSIDER AAEI AS ONE UNIT. THE ARRANGEMENT IS AS FOLLOWS.
M T R L A A E I
THE ABOVE 5 ITEMS CAN BE ARRANGED IN 5! WAYS AND AAEI CAN BE ARRANGED AMONG THEMSELVES IN 4!/2! WAYS.
NUMBER OF REQUIRED WAYS OF ARRANGING THE ABOVE LETTERS = 5! * 4!/2!
= (120 * 24)/2 = 1440 WAYS.
Answer: Option B. -> 600
TWO MEN, THREE WOMEN AND ONE CHILD CAN BE SELECTED IN ⁴C₂ * ⁶C₃ * ⁵C₁ WAYS
= (4 * 3)/(2 * 1) * (6 * 5 * 4)/(3 * 2) * 5
= 600 WAYS.
TWO MEN, THREE WOMEN AND ONE CHILD CAN BE SELECTED IN ⁴C₂ * ⁶C₃ * ⁵C₁ WAYS
= (4 * 3)/(2 * 1) * (6 * 5 * 4)/(3 * 2) * 5
= 600 WAYS.
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