Permutations and Combinations

A. 15
B. 36
C. 48
D. 60
Explanation:
Case I if lady sits on the reserved seat, then 2 men can occupy seats from 4 vacant seats in = 4P2 = 4×3 = 12ways
Case II if lady does not sit on reversed seat, then I. Woman can occupy a seat from four seats in 4 ways. I. man can occupy a seat from 3 seats in 3 ways, also I. man left can occupy a seat from remaining two seats in 2 ways.
Therefore, Total ways = 4x3x2 = 24ways
From case I and case II
Total number of ways = 12+24 = 36

A. 36
B. 81
C. 91
D. 116
Explanation:
When the dice are rolled, the number of possible outcomes = 63 = 216.
Number of possible outcomes in which 2 does not appear on any dice = 53 = 125.
Therefore, Number of possible outcomes in which at least one dice shows 2 = 216- 125 = 91.

A. 12
B. 48
C. 36
D. 42
Explanation:
Married couples = MF MF MF MF
AB CD EF GH
Possible teams = AD CB EB GB
AF CF ED GD
AH CH EH GF S
Since one male can be paired with 3 other female, Total teams = 4×3 = 12.
Team AD can play only with CB,CF,CH,EB,EH,GB,GF(7 teams )
Team AD cannot play with AF, AH, ED and GD
The same will apply with all teams, So number of total matches = 12×7 = 84.
But every match includes 2 teams, so the actual number of matches = 84/2 = 42.

A. 19
B. 120
C. 624
D. 1024
Explanation:
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.

A. 120
B. 240
C. 360
D. 480
Explanation:
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.

A. 5
B. 6
C. 7
D. 8
Explanation:
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.

A. 20
B. 40
C. 512
D. 1024
Each question can be answered in 2 ways.
10 Questions can be answered = 210= 1024 ways.

A. 29
B. 32
C. 55
D. 30
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

A. ²²C₁₀
B. ²²C₁₀ + 1
C. ²²C₉ + ¹⁰C₁
D. ²²C₁₀ – 1
Explanation:
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

A. ¹²C₅ * 10
B. ¹²C₇ * 10
C. ¹²C₇ * ¹⁰C₅
D. 12 * 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