09/08/2018 9:14 pm
PnC , Probability
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TG Team 10/08/2018 12:18 pm
When both ( chocolates and boxes ) identical make cases :
Distributing 1 , 2, 3 ,4 , ......., 9 , 10 chocolates we are left with 57 - 55 = 2 chocolates .
1. ( 1 , 2, 3, 4, 5, 6, 7, 8, 9 + 1 , 10 + 1 )
2. ( 1 , 2, 3, 4, 5 , 6, 7 ,8 . 9+2 , 10)
3. ( 1 , 2,3, 4, 5 , 6, 7 , 8 , 9 , 10+2 )
3 ways .
0
12/08/2018 2:38 pm
how many pairs of x and y are possible if
x2-5y2=1232
TG Team 12/08/2018 4:11 pm
Every perfect square can be written as 5k, 5k+1 or 5k + 4 form.
If x2 is in the form of 5k then x2 – 5y2 will be a multiple of 5.
If x2 is in the form of 5k+1 then x2– 5y2 will be 1 more than a multiple of 5.
If If x2 is in the form of 5k+4 then x2– 5y2 will be 4 more than a multiple of 5.
Now we can see 1232 gives 2 as a remainder when divided by 5,
Hence, x2 – 5y2 = 1232 will have no integer solution.
0
17/08/2018 12:45 am
Four circular tables are arranged such that their centers from a square. If each table can accommodate 4 persons, in how many ways can 16 people be selected across such an arrangement?
Sir, I have a doubt in this question. I agree that for the 1st person, all the tables are alike so he can select any one table in 1 way. After that, why are there 4 different ways for him to be seated at that table? Aren't the chairs on that particular table too alike for him? So shouldn't the ways of selecting the chair be 1 instead of 4? Please clarify.
0
17/08/2018 3:00 am
A box contains 5 chips, numbered from 1,2,3,4 and 5. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds 4. What is the probability that 3 draws are required?
Sir, can you please explain how did we come to the marked part in the above image? I understood the explanation written before it but couldn't understand how did we come to these values.
TG Team 17/08/2018 12:38 pm
4 cases :
( 1 , 2 ) , ( 2 ,1) , (1,3) and (3,1)
probability of getting 1 is : 1/5 and probability of getting 2 from the remaining 4 chips is 1/4
Hence , 1/4 x 1/5 = 1/20
Each of these 4 cases has 1/4 x 1/5 chance . Hence , (1/20) x 4 = 1/5
0
18/08/2018 12:16 pm
TG Team 18/08/2018 1:23 pm
Except the 7 chairs occupied by friends there will be 28 chairs and space between those chairs = 29
Select 7 spaces out of 29 in 29C7ways arrange them in 29C7 x 7!
Hence , 29P7 ways .
0
30/08/2018 1:26 pm
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.
AmitK 30/08/2018 2:57 pm
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.
0
01/09/2018 6:11 pm
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?
TG.Raman 01/09/2018 10:37 pm
_ _ _ _
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 .
0
15/09/2018 10:35 pm
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?
TG Team 16/09/2018 12:21 am
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 .
TG.Kamal.Lohia 18/07/2019 3:27 pm
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. 🙂
0
17/09/2018 3:58 pm
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?
TG Team 17/09/2018 7:36 pm
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 .
0
18/09/2018 5:17 pm
Find the number of ways in which 14 identical balls can be divided into 3 groups?
TG Team 19/09/2018 6:30 am
(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.
Shubhra garg 19/09/2018 1:31 pm
sir i couldnt understand the unordered solution part,why did we subtract 3*6 and then divide by 6 ?
TG Team 19/09/2018 2:23 pm
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 .
0
28/09/2018 6:03 pm
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
TG Team 28/09/2018 6:19 pm
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.
0
06/10/2018 8:31 pm
The number of ways of distributing 20 fruits among 5 people, so that no one receives less than 3 fruits is : __
TG Team 07/10/2018 1:23 pm
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 .
0
10/10/2018 3:05 pm
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.
0
10/10/2018 11:23 pm
Utkarsh Garg 10/10/2018 11:35 pm
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
0
12/10/2018 4:08 pm
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?