Permutation and Combination-Test 1

first two places can be filled in 5 and 4 ways respectively so, total number of 3 digit number =5*4*1 = 20

Vowels = E, E and A. They can be arranged in 3!/2! Ways
so total ways = 6!*(3!/2!) = 2160

Zero or more bananas can be selected in 4 + 1 = 5 ways (0 orange, 1 orange, 2 orange, 3 orange and 4 orange)
similarly apples can be selected in 7 +1 = 8 ways
and mangoes in 6 +1 = 7 ways
so total number of ways = 5*8*7 = 280
but we included a case of 0 orange, 0 apple and 0 mangoes, so we have to subtract this, so 280– 1 = 279 ways

From 15 points number of lines formed = 15c2
6 points are collinear, number of lines formed by these = 6c2
So total lines = 15c2 – 6c2 + 1 = 91

4 Indians can be seated together in 4! Ways, similarly for Africans and Japanese in 5! and 7!
respectively. So total ways = 4! 5! 7! 3!

First Indians can be seated along the circular table in 4! Ways and now Americans can be
seated in 5! Ways. So 4! 5! Ways

Every chess match can have three result i.e. win, loss and draw
so now of ways = 3*3*3*3 = 81 ways

6c2*5c3 + 6c3*5c2 + 6c4*5c1 + 6c5

Total cards = 52
Drawn cards(Heart) = 2
Present total cards = total cards-drawn cards =52-2=50
Remaining Card 13-2 = 11
Probability = 11/50

Probability = (3/5 * 5/8)/([3/5 * 5/8] + [2/5 * 1/3]) = 45/61

P(A) = 20c2 / 30c2, P(B) = 22c2 / 30c2
P(A∩B) = 15c2 / 30c2
P(A∪B) = P(A) + P(B) – P(A∩B) => (20c2/30c2)+(22c2/30c2)-(15c2/30c2)=316/435

Five persons is to be chosen from a group of 10 people = 10C5 = 252
Couple Serve together = 8C3 * 2C2 = 56
Couple does not serve = 8C5 = 56
Probability = 102/252 = 51/126

Man only = 8C2 = 14
Probability of selecting no woman = 14/91
Probability of selecting atleast one woman = 1 – 14/91 = 77/91

Total probability = 6C2 = 15
Probability that one is Banana and another one is orange = 3C1 * 3C1 = 9
probability = 9/15 = 3/5

Total Balls = 15
Probability = 6c3 / 15c3 = 4/91

Total Balls = 15
Probability = 4c2 * 3c1/ 15c3 = 18/455

Total Balls = 15
Probability = 11c4/15c4 = 22/91
One is black = 1 – 22/91 = 69/91

Probability both are Pink = 1/15C2
Probability both are Green = 3/15C2
Required Probability = 4/15C2 = 4/105

NUMBER is 6 letters. We have 4 places where letters are to be placed. For first letter there are 6 choices, since repetition is not allowed, for second, third and fourth letter also we have 5, 4, and 3 choices resp., so total of 6*5*4*3 ways = 360 ways.

ALLIGATION contains 10 letters, so total 10! ways. There are 2 As, 2 Ls, 2 Is
So 10!/(2!*2!*2!)

Take vowels in a box together as one – IIU, M, N, M, M
So there are 5 that to be placed for this 5!, now 3 Ms, so 5!/3!, so arrangement of vowels inside box gives
3!/2!
So total = 5!/3! * 3!/2!

Total ways = 8C4*5C3

A number is divisible by 4 when its last two digits are divisible by 4
For this the numbers should have their last two digits as 00, 04, 08, 12, 16, … 96
By the formula, an = a + (n-1)d
96 = 0 + (n-1)*4
n = 25
so there are 25 choices for last 2 digits and 9 choices (1-9) for the 1st digit
so total 9*25=225

0 cannot be placed at first digit to make it a 3 digit number.
3 cases:
Case 1: 2 is placed at first place
1 choice for the first place, 9 choices each for the 2nd and 3rd digit (0-9 except 2)
So numbers = 1*9*9 = 81
Case 2: 2 is placed at second place
8 choices for the first place (1-9 except 2), 1 choice for the 2nd digit and 9 choices for the 3rd digit (0-9
except 2)
So numbers = 8*1*9 = 72
Case 3: 2 is placed at third place
8 choices for the first place (1-9 except 2), 9 choices for the 2nd digit (0-9 except 2) and 1 choice for the
3rd digit

So numbers = 8*9*1 = 72
So total numbers = 81+72+72 = 225

Total 15 people, and a particular woman is to be taken to form a group of 5, so choice is to be done from
14 people of 4 people
Ways are 14C4.