Pls help
No of ways in which 3nos. In a.p can be selected from 1,2,3....n
@macc6 said:Kindly helpThe no of ordered triplets of +ve inegers which are solution of the equation~ x+y+z=100
you can apply direct formula i.e. number of integral solutions
but here we need only +ve integers
x = x' + 1
y = y' + 1
z = z' + 1
x' + 1 + y' + 1 + z' + 1 = 100
=> x' + y' + z' = 97
now apply integral solution formula i.e. (97 + 2)C2
but here we need only +ve integers
x = x' + 1
y = y' + 1
z = z' + 1
x' + 1 + y' + 1 + z' + 1 = 100
=> x' + y' + z' = 97
now apply integral solution formula i.e. (97 + 2)C2
Help
Using all Sum of all the numbers that can be formed by 2,3,3,4,4,4
Hi Gurus , these question appeared in IIFT 2012 kindly help me to find the solutions, Thanks a ton in advance !
Q1) The answer sheet of 5 engineering students can be checked by any one of 9 professors.What is the probability is that all the 5 answer sheets are checked by exactly 2 professors ?
A)20/2187 B)40/2187 C)40/729 D)None of the above
Q2) Ashish is studying late into night and is hungry.He opens his mother's snack cupboard without switching on the lights,knowing that his mother has kept 10 packets of chips and biscuits in the cupboard.He pulls out 3 packets from the cupboard , and all of them turn out to be chips .What is the probability that the snack cupboard contains 1 packet of biscuit and 9 packets of chips?
A)6/55 B)12/73 C)14/55 D)7/50
@Alex.
Question 1
Favourable cases :
Select any two teachers out of the 9 = 9C2 ways
Now no. of ways in which 5 questions may be allocated to these 2 teachers = 2^5
But in these 2^5 cases, there will be two cases such that all five questions are allotted to a single teacher.
So total ways of allotting questions to exactly 2 teachers is = 2^5 - 2
Total favourable cases = 9C2 . (2^5 - 2) = 36.30
Total Possible cases = 9^5
Ans = 36.30/9^5 = 40/2187
Question 1
Favourable cases :
Select any two teachers out of the 9 = 9C2 ways
Now no. of ways in which 5 questions may be allocated to these 2 teachers = 2^5
But in these 2^5 cases, there will be two cases such that all five questions are allotted to a single teacher.
So total ways of allotting questions to exactly 2 teachers is = 2^5 - 2
Total favourable cases = 9C2 . (2^5 - 2) = 36.30
Total Possible cases = 9^5
Ans = 36.30/9^5 = 40/2187
@Alex.
Question 2
Total cases possible for drawing three chips - 10C3+9C3+8C3+7C3+6C3+5C3+4C3+3C3 = 120+84+56+35+20+10+4+1 = 330
Favourable case - 9C3 = 84
Ans = 84/330 = 14/55
Question 2
Total cases possible for drawing three chips - 10C3+9C3+8C3+7C3+6C3+5C3+4C3+3C3 = 120+84+56+35+20+10+4+1 = 330
Favourable case - 9C3 = 84
Ans = 84/330 = 14/55
@pendyal said:first choose a rock song and a carnatic song.this can be done in 5 ways and 6 ways respectively.so we can choose a rock song and a carnatic song together in 30 ways.now of the remaining 12 songs,each song can either be in the album or not be in the album.that is there are 2 ways for each of the 12 songs.so these 12 songs can be chosen in 2*2*2....*2 (i.e. 2^12).the total no. of ways becomes 30*(2^12).bye..
guys, years ago i made this post....problem is, it is wrong. using the abov logic we'll be counting some of the albums multiple times.
the correct way of solving this wud be (2^5-1)*(2^6-1)*2^3
2^5-1 becoz there are two options for each rock song. hence total of 2^5 options. but one of these wud be no rock song inluded. hence subtracting 1. same reasoning for carnatic. since there is no restriction on the minimum number of indie pop songs, i dont subtact 1.
P.S. been so long since i posted on this thread...some users liking the above post brought the error to my notice.
Probability and combinatorics
Permutations and combinations. Using combinatorics to solve questions in probability. https://www.khanacademy.org/math/probability
Permutations and combinations. Using combinatorics to solve questions in probability. https://www.khanacademy.org/math/probability
Probability and combinatorics
Permutations and combinations. Using combinatorics to solve questions in probability. https://www.khanacademy.org/math/probability
Permutations and combinations. Using combinatorics to solve questions in probability. https://www.khanacademy.org/math/probability
Probability and combinatorics
Permutations and combinations. Using combinatorics to solve questions in probability. https://www.khanacademy.org/math/probability
Permutations and combinations. Using combinatorics to solve questions in probability. https://www.khanacademy.org/math/probability
@vishal.das said:@visionIIM-ACL@deepu Is above post ad ?
Sir,
Khan academy is one of the most renowned website by a MIT Haward grad. He even gave a TED talk which was appreciated by Bill Gates.
Read this: http://en.wikipedia.org/wiki/Khan_Academy
The latest number issued by the Delhi Motor vehicle Registrtion authority is DL-5S 2234 . If all the numbers and alphabets before this have been used up then find how many vehicles have registration number starting with DL-5 ?
- none of these
- 172227
- 192234
- 192225
- 182216
0 voters
Arranging the letters of the word DISCUSSION what is the probability that all vowels are not together?
The solution for this problem will 10!/3!2! because the word contains 10 letter and s is reapeated 3 times and i is repeated 2 time.
A 10-member student leadership committee consists of juniors and seniors. There are 4 junior and 6 senior students. Exactly 6 students will be selected from this group to attend a national convention. What is the probability that at least 3 seniors are selected for the committee?
I am unable to solve this question using the below method (but have been able to solve it using the other methods such as using Combinations Formula)
My approach which is going wrong-->Since atleast 3 have to be Seniors,
Probability of Seniors 3- Juniors 3 --> (3/6*3/4)
Probability of Seniors 4- Juniors 2 --> (4/6*2/4)
Probability of Seniors 5- Juniors 1 --> (5/6*1/4)
Probability of Seniors 6- Juniors 0 --> (6/6*0/4)
Net Proobability..Summing all of the above 9/24+8/24+5/24+0=22/24=11/12
But the answer is 13/14
Please let me know where am I going wrong in the above solution
Is there any post regarding distribution of n balls to m boxes when(i)both balls and boxes are different.(ii)balls are similar but boxes are different.(iii)balls are different but boxes are similiar.(iv)both balls and boxes are similar.HELP PUYS!!!!!
hello everyone...can someone tell me good books for quant?
How many ways 30 identical chocolates can be distributed to 4 persons such that all are getting odd number of chocolates?
PS: I don't know the correct answer.
- 216
- 288
- 560
- 480
0 voters
