Permutation and com...

# Permutation and combination

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1. Number of selections of atleast 10 articles from 19 different articles is.
A) 2^19      B) 2^18       C) 2^19 - 1        D) 2^18 - 1      E) None

55 Answers
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Hello sir , kindly share the solution of the following problem-

A round table with 5 chairs is to be decorated. Each chair can be painted with one of  5  colors available. In how many different ways the  chairs can be painted?

This question was asked in copycat 03 , and I did not understand the solution given there.

First chair can be painted in 5 ways .
for second chair , again we have 5 possibilities.
same for 3rd , 4th and 5th chair .

Hence total : 5^5 ways .

Now when all chairs are coloured with the same colour there is just one way and there are 5 colours available so 5 ways . ( They are counted just once )

cases where all the chairs are not coloured with the same colour are counted 5 times hence ,

(5^5 - 5)/5 + 5 = 629 cases.

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How many 4-digit numbers can be formed which are divisible by 4 , using the first 8 whole numbers , if repetition of digits is not allowed?

_ _ _ _

Last two digits should be divisible by 4.

List down all the possibilities for last two digits

=> 04, 12, 16, 20, 24,32,36,40,44,52,56,60,64,72,76.

Now, 44 is not possible since it contains repetitive digit.

Now make two cases:
Case - 1:  The number which contains 0 in either of the last two digits.

=> 04,20,40,60 ( 4 numbers )

So there are total 6 × 5 × 4  =120 Numbers .

Case - 2: The numbers  which does not  contain 0 in the last two digits.

12,16,24,32,36,52,56,64,72,76  (10 numbers )

So ,  5 × 5 × 10 = 250 Numbers .

Hence , 120 + 250=370 Numbers .

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In how many different ways 8 different balls can be placed into 3 identical boxes if (a) blank box is not permitted, (b) blank box is permitted?

Hello Samyak ,

When blank box is not permitted - use inclusion exclusion .

(3^8 - 3C1 × 2^8 + 3C2 × 1^8)/3! .

When blank box is permitted :

(3^8 - 3 × 1)/3! + 1 .

What is the explanation for the (3^8 - 3 × 1)/3! + 1?

When all the 3 boxes had been different, then number of arrangements were 3^8. Right?
But if the three boxes are same then arrangements within 3 different places i.e. 3! becomes 1. So the previous answer needs to divided by 3!.
Now there is one more thing that exactly one case (8, 0, 0) has been counted 3 times in 3^8 and all others 3! = 6 times.
So taking all these points in consideration, we get the final answer expression as
(3^8 - 3)/3! + 1 OR (3^8 +3)/3!

I hope it is clear now. 🙂

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A,S, and P go to a seaside town  and they decide to visit different tourist locations. After breakfast each of them boards a different tourist vehicle from the nearest bus stop. After 3 hours, S , who had gone to a beach , calls on the mobile of P and claims that he has observed a shark in the waters. P learns from the local guide that at that time of the year , only 8 sea creatures(including shark) are observable and the probability of observing any creature is equal. However , A and P later recall during their discussion that S has a reputation for not telling the truth 5 out of 6 times. What is the probability that S actually observed a shark in the waters?

S sees a shark and reports a shark : 1/8 x 1/6

Now, he could have not seen a shark and reported a shark : 5/6 x 7/8

so probability = (1/6 x 1/8)/(1/6 x 1/8 + 5/6 x 7/8) = 1/36 .

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Find the number of ways in which 14 identical balls can be divided into 3 groups?

(x +1) + ( y +1) + ( z +1) = 14

x + y + z = 11

Positive Integral Ordered Solution : (11+3 -1)C( 3-1) = 13C2 =  78

Unorded Solution: (78 - 3 × 6)/6 + 6 = 16 ways.

sir i couldnt understand the unordered solution part,why did we subtract 3*6 and then divide by 6 ?

Hi Shubhra,

Balls are identical so we need unordered solution .

When two groups get similar number of balls

2x + y = 11

x can vary from 0 to 5 i.e 6 cases . ( 0 , 0 , 11)  , ( 1, 1, 9) , ...., ( 5,5,1)

ordering x , x, y we get 3!/2! i.e  3 cases , so total 3 x 6 cases of x , x, y type.

removing these cases we are left with : 78 - 18 = 60 cases .

Each case is repeated 3! times in (x ,y,z type ) so divide 60 by 6 .

Total : 10 + 6 => 16 unordered solution .

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In how many ways can 5 identical balls be placed in 5 identical boxes such that there can be any number of balls in any of the box

Just count manually

5 , 0, 0 , 0, 0
4, 1, 0 , 0, 0
3, 2, 0, 0, 0
3, 1, 1, 0, 0
2, 2, 1, 0, 0
2, 1, 1, 1, 0
1, 1, 1, 1, 1

7 ways.

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The number of ways of distributing 20 fruits among 5 people, so that no one receives less than 3 fruits is : __

a + b + c + d + e = 20

to make at least 3

( a' + 3) + ( b' + 3) + ( c' + 3) + ( d' + 3) + ( e' + 3) = 20

a' + b ' + c' + d' + e' = 5

Total : (5+5 -1)C(5-1) = 9C4  whole number solutions .

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How many different paths are there on a regular 8 × 8 chessboard from the lower left corner to the upper right corner? The path is always going along the edges of the unit squares to the right, up or down but not left. Also no passage should be retraced on a path.

What are the options

Answer is 21!/(7!)^3 ?

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Urn 1: 5 white, 7 black
Urn 2: 7 white, 8 black.

We move a ball from urn 1 to urn 2. It had a 5/12 chance of being white and 7/12 chance of being black. The second urn is now

(5/12): 8 white, 8 black
(7/12): 7 white, 9 black

The chance if drawing a white at this point is either 8/16 or 7/16. Weighting it by the chance if being in either case
5/12*8/16 + 7/12*7/16

89/192

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In how many ways two red balls, one green ball and one blue ball can be put in three boxes of different sizes if order of putting the balls doesn’t matter?

54 or 81?

a+b+c=2
a+b+c=1
a+b+c=1

multiply all solutions=> 4C2 x 3C1 x 3C1 = 54

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shortest (1) and tallest (8) women are fixed:
XXX8
1XXX

2 and 7 have two options, so total 4 patterns:

1:
2xx8
1xx7

|2468| |2568|
|1357| |1347|

2.
2x78
1xxx

|2478| |2578| |2678|
|1356| |1346| |1345|

3.
xxx8
12x7

|4568| |3468| |3568|
|1237| |1257| |1247|

4.
xx78
12xx

|3478| |3678| |3578| |5678| |4578| |4678|
|1256| |1245| |1246| |1234| |1236| |1235|

14 options.

Answer: B.

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Solution for question 24 please.

TTT | AAAHG

_ A _ A _ A _ H _ G_:  Arrange these letters in 5!/3! i.e 20 ways

space among these letters : 6 .

Select any three spaces in 6c3 ways

Hence , 6C3 x 20 = 400 ways .

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solution for questions 5 & 6 please.

6th .

other than these 3, there are 5 persons. between these persons there are 6 gaps. the chosen 3 can be anywhere in these 6 gaps , So 6C3 = 20 ways .

5th

ABCD

B > C > D.

A can take 9 values . ( 1, 2, 3, 4,...., 9)
For D = 1 , select B and C from ( 2 , 3, 4, ... 9) in 8C2 ways .
For D = 3 , select B and C from ( 4 , 5 , ...., 9) in 6C2 ways
Similarly for D =5 , 7 , 9 there are 4C2 , 2C2 and 0 ways respectively .

Hence total : 9 x ( 8C2 + 6C2 + 4C2 + 2C2 ) = 50 x 9 = 450 Numbers .

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please provide the solution for question 30 part (iii) and question 36 (with explanation)

Solution to the question 36 .

Let box 1 contains black, others white. Similarly box 2 black, others white..so on. Therefore, case of one box containing black has 6 possibilities.
Similarly 1 and 2 contain black , rest white. 2 and 3 contain black, rest white..and so on. So 5 possibilities

Similarly for 3, 4,5 and all 6 boxes containing black balls.
So total
6 + 5+4+3+2+1 = 21 possibilities

30th part (iii)

three cases will be formed with atleast 2 girls

(2G,2B)   (3G,1B)  (4G,0B)

5C2*4C2 +5C3*4C1 +5C2*4C0

10*6+10*4+5=105 WAYS

Alternate Approach :

Total : 9C4 = 126

remove cases when there is no girl and 1 girl.

So , 126 - 4C4 - 5c1 x 4C3 = 126 - 1 - 20  = 105

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