# Permutations & Combinations - Questions & DiscussionsQuantitative Report

.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 ...
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Arranging the letters of the word DISCUSSION what is the probability that all vowels are not together?​
8 users have answered this question.
Arranging the letters of the word DISCUSSION what is the probability that all vowels are not together?​
1 user has answered this question.
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Probability and combinatorics
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
Probability and combinatorics
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...
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@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
@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...
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 ?...
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
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 =...
@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??
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_ RED _ WHITE _
g1 + g2 + g3 = 6 (where g2 > 0)
=> number of solutions = 7C2 = 21
and in 2! ways red and white arrange themselves
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)
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!
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!
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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
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
bhai 52*36*22*10 hoga na. to choose a suit also.
13x12x11x10
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...
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.......
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@pendyal @keviv @GauravShah @maddy2807 @naarto @naga25french @Aizen @Estallar12 @chillfactor
Can Anyone of you help me out with this one??
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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...
@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