4 balls are to be put in 5 boxes.in how many ways can this be done if
a)Balls are similar and boxes are different
b)balls are different and boxes are similar
c)both boxes and ...
10
10
.Can someone help me out with the following problems with some clear and simple approcah to these qustions
4 balls are to be put in 5 boxes.in how many ways can this be done if
a)Balls are similar and boxes are different
b)balls are different and boxes are similar
c)both boxes and ...
4 balls are to be put in 5 boxes.in how many ways can this be done if
a)Balls are similar and boxes are different
b)balls are different and boxes are similar
c)both boxes and ...
Arranging the letters of the word DISCUSSION what is the probability that all vowels are not together?
Arranging the letters of the word DISCUSSION what is the probability that all vowels are not together?
1 user has answered this question.
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
@visionIIM-ACL @deepu Is above post ad ?
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
@pendyal sir ____/\____
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 th...
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 th...
1
@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
@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...
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...
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 ?...
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 ?...
HelpUsing all Sum of all the numbers that can be formed by 2,3,3,4,4,4
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
Pls helpNo of ways in which 3nos. In a.p can be selected from 1,2,3....n
@mayurdhingra thnxx a lot...
@macc6
x + y + z = 100;
0 < x<99
0< y < 99
0<99
x , y, and z can only be from 1 to 98
Because if the other two nos, are both 1, the value of the third is 98.
if x = 1, y could only be 1 to 98 (98 choices)
if x = 2, y could only be 1 to 97 (97 choices)
......
If x =...
x + y + z = 100;
0 < x<99
0< y < 99
0<99
x , y, and z can only be from 1 to 98
Because if the other two nos, are both 1, the value of the third is 98.
if x = 1, y could only be 1 to 98 (98 choices)
if x = 2, y could only be 1 to 97 (97 choices)
......
If x =...
@macc6 take all cases and subtract those in which red and white come together.So, total cases = 8!Cases in which red and white come together = 7!*2!So, required answer = 8! - 7!*2!
Kindly helpThe no of ordered triplets of +ve inegers which are solution of the equation~ x+y+z=100
@jain4444 Can you tell me what is g1,g2,g3??
_ RED _ WHITE _
g1 + g2 + g3 = 6 (where g2 > 0)
=> number of solutions = 7C2 = 21
and in 2! ways red and white arrange themselves
required answer = 21*6!*2!
g1 + g2 + g3 = 6 (where g2 > 0)
=> number of solutions = 7C2 = 21
and in 2! ways red and white arrange themselves
required answer = 21*6!*2!
8 balls of different colours are arranged in a row.Condition- red and white may not come together.
Agreed. There are two ways to reach the solution-
i) C(52,1)*C(36,1)*C(22,1)*C(10,1)/4!ii) C(13,1)*C(12,1)*C(11,1)*C(10,1)
i) C(52,1)*C(36,1)*C(22,1)*C(10,1)/4!ii) C(13,1)*C(12,1)*C(11,1)*C(10,1)
Are they ?? Then might be I was calculating your answer incorrectly ...
Both answers are same buddy :)
Are you sure?
Check this:
You can select 4 different numbers in 13C4 ways
Now you have to assign 4 different suits to these 4 numbers: which you can do in 4! ways
Total no. of ways: 13C4 * 4!
Check this:
You can select 4 different numbers in 13C4 ways
Now you have to assign 4 different suits to these 4 numbers: which you can do in 4! ways
Total no. of ways: 13C4 * 4!
1st select any card = 52C1
now we have left with 52 - 13 - 3 = 36 cards
2nd card = 36C1 in same way other 2 in 22C1 and 10C1
total ways = 52C1*36C1*22C1*10C1
but here order doesn't matter so , we have to divide it by 4!
required answer will be = 52C1*36C1*22C1*10C1/4!
now we have left with 52 - 13 - 3 = 36 cards
2nd card = 36C1 in same way other 2 in 22C1 and 10C1
total ways = 52C1*36C1*22C1*10C1
but here order doesn't matter so , we have to divide it by 4!
required answer will be = 52C1*36C1*22C1*10C1/4!
2
bhai my doubt is shud me choose suits also?
bhai 4 suits bole to 13,13,13,13
choose 1 from any1 - 13c1
ab baki sab se 1 option kam, i.e 12c1, similarly
13c1*12c1*11c1*10c1
choose 1 from any1 - 13c1
ab baki sab se 1 option kam, i.e 12c1, similarly
13c1*12c1*11c1*10c1
13 cards in each suit.
Now, if you select any one of the 13 cards of say spades ; u'll be left with only 12 options to choose from the next set of suit & so on.........
Arun sharma ka qstn hain bhai ; kiya hua hain boht baar.
baaki dekh lo ....letz see what @krum & @rkshtsurana says
Now, if you select any one of the 13 cards of say spades ; u'll be left with only 12 options to choose from the next set of suit & so on.........
Arun sharma ka qstn hain bhai ; kiya hua hain boht baar.
baaki dekh lo ....letz see what @krum & @rkshtsurana says
bhai 52*36*22*10 hoga na. to choose a suit also.
13x12x11x10
start posting on quant thread if you want an instant reply.
start posting on quant thread if you want an instant reply.
Kindly help me with this one:
In how many ways we can select four cards of different suits & different denominations from a deck of 52 cards??
Confusion is between : 52 X 36 X 22 X 10 and 13 X 12 X 11 X 10
Which one is correct ???
@LeoN88 Can Anyone of yo...
In how many ways we can select four cards of different suits & different denominations from a deck of 52 cards??
Confusion is between : 52 X 36 X 22 X 10 and 13 X 12 X 11 X 10
Which one is correct ???
@LeoN88 Can Anyone of yo...
thnx dude!!
In 2nd case repetitions are taking place eg. You choose B1 & G1 first then selecting B4,G2 OK Then in another case you choose B1 & G2 first then selecting B4 & G1.......
1
@pendyal @keviv @GauravShah @maddy2807 @naarto @naga25french @Aizen @Estallar12 @chillfactor
Can Anyone of you help me out with this one??
Can Anyone of you help me out with this one??
1
This is a very easy question & i already know the solution but i am confused about another approach(May be my concepts are not clear)..ďťż
Q. There are 5 boys & 6 girls. A committee of 4 is to be selected so that it must consist of at least 1 boy & 1 girl??
Conventional way to solve this is...
Q. There are 5 boys & 6 girls. A committee of 4 is to be selected so that it must consist of at least 1 boy & 1 girl??
Conventional way to solve this is...
@somu6211 dost so u mean ans is option b right ?
@@ankur_tiger options for that question 1)0 2)1 3)3 4)9 I think 1 is the right answer
@maddy2807
Ans shud be 1 as 52*9mod 17 gives 1 as remainder.
Ans shud be 1 as 52*9mod 17 gives 1 as remainder.
36/71?
Two students play a game based on the total roll of two standard dice. Student A says that a 12 will be rolled first. Student B says that two consecutive 7s will be rolled first. The students keep rolling until one of them wins. What is the probability that A will win?
vowel can be arranged in 5c2 * 2 ! = 20 ways
consonant can be same or different .. so 17^2 = 289
so in total 5780 ways
consonant can be same or different .. so 17^2 = 289
so in total 5780 ways
1
5440?
A few number of four-letter words are formed by using 17 consonants and 5 vowels. How many words out of them will have 2 different vowels in the middle and a consonant at each end?
aisa bhi mat bolna :P
@naarto question wahi hai jo aap bol rahe ho..yeh question kaha aya hai pata hai mujhe..bt shayad bola nahi jasakta hai abhi uska source.. :P
1
the question shld have been the remainder when no of ways of drawing the cards in such way divided by 17. Bas aisa kahi dekha lag raha mujhe isliye
Bas aisa kahi dekha lag raha mujhe isliye
bhai Not at all sure for the ans... I hv taken general cases and added up the Numbers.... took ace as 1 and king as 13... correct me if i m wrng?
kaise? can u enlighten!
6?
In a pack of cards of 52,if a card is randomly picked and its number is noted down and place the card in the pack.This process is repeated 9 times .what is the remainder when the sum is divided by 17
@srikanthp09 ans 540 is possible only if we assume that 3 boys are identical.. though it sounds meaningless but thats how the solution is coming.
@didroo you have to select a pair of shoe so that can be done only in 10 ways as there are 10 pairs available and once a pair is selected you can select one shoe from remaining 18 shoes i.e. 18 ways to do it. that's why 10*18
We play the coin tossing game in which if tosses match, I get both coins; if they differ, you get both. If you have m coins and I have n, what is the expected length of the game (i.e., the number of tosses until one of us is wiped out)?
@swap747 yup u are correct..then i guess the answer is 13/18. (26/36)
@nik25 dude....its not 55...u r correct till a+b =7 but for a+b= 8 its not 7, rather its 5 and it decreases from then i.e for a+b= 12...its a=6 and b=6 so 1way. in any case maximum number of sets possible are 36. so even if u get 55, some have to be redundant.
@swap747 a+b can take values from 2 to 12, but N will be a multiple only when a+b is 6,7,8,9,10,11,12. a+b=2 --1 way, a+b=3 --2 ways in this way total will be 55 (1+2+3+...11)
@nik 25 posted the question, hoping someone wud come up with a proper explanation to that
@nik25 dude, one thing i am not getting is, 55? the total number of set of values of a,b are 36. for it to be a multiple it has to >=6 and <=12, so required set of vales that satisfies are 26...so it shud be 13/18.
earlier, i did a blunder, was trying to find divisors instead of multiples ......
earlier, i did a blunder, was trying to find divisors instead of multiples ......