1.There are 8 orators A,B,C,D,E,F,G,H. in how many ways can the arrangements be made so that A always comes before B and B before C?
a-8!/3!
b-8!/6!
please see the solution in the number system thread.
please see the solution in the number system thread.
1.There are 8 orators A,B,C,D,E,F,G,H. in how many ways can the arrangements be made so that A always comes before B and B before C?
Solution:
8 ppl can be arranged in 8! ways.
Considering 'ABC' form a group we get 2 Groups of 3 members and 5 members.
So, the answer id 8!/3!5! ways.
3! = 3*2*1 = 6
So, we can write the Answer as 8!/6! ways (b)
how many natural nos havn at most 5 digits have sum of their digits as at most 5?
pls do give a detailed sol
how many natural nos havn at most 5 digits have sum of their digits as at most 5?
pls do give a detailed sol
a + b + c + d + e = 5
Now each one can vary from 0 to 5.
(n+r-1) C (r-1) = 9C4
Similarly if Sum = 4 ==>> 8C4
Similarly if Sum = 3 ==>> 7C4
Similarly if Sum = 2 ==>> 6C4
Similarly if Sum = 1 ==>> 5C4
So the required answer = 9C4 + 8C4 + 7C4 + 6C4 + 5C4
Sayonara
@Stalworth- How can the sum of 1 be expressed in 5 ways. here, only a can be 1, Otherwise it does not remain a 5 digit number anymore.
Sum 1= 1 way
sum 2=4c1 ways
sum 3=4c2+4c1x2
sum 4=4c3+4c1+3x4c2+4c1x3c1
and so on...
i ve real problem with this- Find the number of positive integral solution of the inequality 3x+y+z this has been solved before.
x=1, y+zpls give a detailed approach
@Stalworth- How can the sum of 1 be expressed in 5 ways. here, only a can be 1, Otherwise it does not remain a 5 digit number anymore.
Sum 1= 1 way
sum 2=4c1 ways
sum 3=4c2+4c1x2
sum 4=4c3+4c1+3x4c2+4c1x3c1
and so on...
How can Sum of 1 can only be expressed in one way .. See below
10000, 1000, 100 ,10 and 1 .. Are they not satisfying the conditions?
Sayonara
In a temple there r 12 pair of shoes. A boy steals 4 shoes.whats t probablty tht atleat one Left-Right pair of matchng shoes is stolen.?
a)4818/5313
b)3856/5313
c)4686/10626
d)6058/10626
ok,MAybe this would be the easiest question out there but i am not able to get funda behind the same:
In how many ways can 32 different books be divided equally amongst
1.4 Boys
2.4 Parcels
How are they different?
samajh nahin aa raha!
ok,MAybe this would be the easiest question out there but i am not able to get funda behind the same:
In how many ways can 32 different books be divided equally amongst
1.4 Boys
2.4 Parcels
How are they different?
samajh nahin aa raha!
4 Boys can well be trated as DIFFERENT things while Parcels as SAME!
Sayonara
In the first case the answer will be (32C
(24C (16C(8C
= (32!)/ (8!)^4
In the second case the parcels are non distinguishable, hence the answer will be [ (32C (24C (16C(8C]/ 4! = (32!)/ {(8!)^4. 4!}
hi ,
Please help me in the following question
1) Find the number of selections that can be made by taking 4 letters from the word INKLING?
2)arrangements for by taking 4 letters for the word above
Thanks
Rajesh
hi ,
Please help me in the following question
1) Find the number of selections that can be made by taking 4 letters from the word INKLING?
2)arrangements for by taking 4 letters for the word above
Thanks
Rajesh
I dont know why this question has been left unslolved... So even though u might have got the answer it could be helpfull someone else who is looking out for solution!!
For first
The word INKLING has 5 distinct letters and 2 letters are repeating...
1.U can select 4 letters from 5 distinct letters in 5c4 ways=5
2.When 2 I`s are repeating, u can arrange them in 1x4c2 ways=6(1x5c2 bcoz u can select `I`I` in one way and remaining 2 letters out of 4 distinct letters can be done in 4c2 ways)
3.When 2 N`s are repeated, u can arrange them in 1x4c2 ways=6
4.the combination IINN can be done in 1way... Remember the arrangement (ININ or NINI or NNII or any other arrangement doesn`t matter)
so 5+6+6+1=18 ways
2) Now second is bit tricky
1)arrange 5 distinct letter(INKLG) This can be done in 5 x 4 x 3 x 2 x1 =120
2)arrange II and NKLG, this can be done in 13/2!x4x3=72
(12/2! bcoz I can be arranged in 4 ways and other I can be arranged in 3 ways.. Dividing by two bcoz we might encounter a sitation of I1I2KLor I2I1KL so we dont want such combunation)
3)arrange NN and IKLG this can be done in 72 ways similar to last case
4)arrange II and NN this can be done in 4!/2!2!=6 ways
so total arrangement of 120+72+72+6=270
I hope I got both of them right
i am in a little trouble with the permutation and combination questions as i am not able to determine whether permutation is used or combination
actually i do have understood the way to work out the questions but i still do not know the basic difference b/w the two
i m understanding permutation is the ways to arrange things
whereas combination is the way for selection
but how to get this logic under use i am not able to understand from last few mocks i am getting the questions of P & C wrong............just of this i do not know how to determine what is the question for
for example:
2 men ,3 women ,4 girls and a boy are present at the meeting where committee of 5 members are to be formed how can it be done so....??
now i want to know
how you understand it to be a question of permutation or combination ?
are there any keywords one must look for...??
i know its a stupid question when cat is just 3 months ahead but i m loosing my grounds because of this in P & C which i beleive can make me to score a little more then i can do otherwise......
plz help 
i am in a little trouble with the permutation and combination questions as i am not able to determine whether permutation is used or combination
actually i do have understood the way to work out the questions but i still do not know the basic difference b/w the two
i m understanding permutation is the ways to arrange things
whereas combination is the way for selection
but how to get this logic under use i am not able to understand from last few mocks i am getting the questions of P & C wrong............just of this i do not know how to determine what is the question for
for example:
2 men ,3 women ,4 girls and a boy are present at the meeting where committee of 5 members are to be formed how can it be done so....??
now i want to know
how you understand it to be a question of permutation or combination ?
are there any keywords one must look for...??
i know its a stupid question when cat is just 3 months ahead but i m loosing my grounds because of this in P & C which i beleive can make me to score a little more then i can do otherwise......
plz help
Well, no question is stupid so you need not worry about that.
Simply speaking, permutations is the number of ways to arrange things in a specific order and combinations is a way of arranging things such that the order is not important.
As in, taking your example,
2 men, 3 women, 4 girls and 1 boy
You need a team of 5 from these 10 people.
Now, without any condition if we have to pick 5, we can do it in 10C5 ways. Remember here that the order is not important.
Lets take another example. Say you have to pull out 2 cards from a pack of cards. Say you want to get an ace of spades and an ace of clubs. How would you do it?
2 cases:
1) The order is important: It is specifically mentioned that the first one be a spade and the second one a club
2) The order isn't important, it can be either a spade first than a club or vice versa.
So, for the first case, as the order is important, you have to consider the outcomes separately:
First card has to be an ace of spades. It can be done in 1/52 ways.
Second one has to be an ace of clubs. It can be done in 1/51 ways.
Total of 1/(51*52) ways.
Now,the second case:
We can get an ace of spades and an ace of clubs in only 1 way. The total ways of pulling out 2 cards from the deck would be 52C2 which is 51*26 ways.
Now, why are the 2 figures different?
As the order isn't important in the second case, it's fine if you get (1S, 1C) or (1C, 1S). So, the possibilities are doubled.
Now, a question for you.
What is the probability of getting a total of 10 if you roll 2 dice
1) of the same colour
2) of different colours
In the above question we are not concerned with arrangement. Its just selection which matters
As shashank Bhai said This can be done in 10C5 ways... As ur just concerned with selecting 5 people out of 10 people!!
Nicely explained by shashank Bhai if u still have confusion u can through these articles...
http://books.google.co.in/books?id=ilhcN9ymMBcC&lpg;=PP1&pg;=PA441#v=onepage&q;&f;=false
Also go through this article - Pand c
Well, no question is stupid so you need not worry about that.
Now, a question for you.
What is the probability of getting a total of 10 if you roll 2 dice
1) of the same colour
2) of different colours
different colours
well to get 10 the cases are (6,4) ,(4,6) & (5,5)
and probability of getting any sum is 1/36
so forgetting 10 we have a prob. of 1/12
also colour(if of same colour) will just reduce the cases of (6,4) ,(4,6) as one
hence prob. will become 1/18
am i nearby.......?
Dear Puy,
I thought that your explaination is right but I just wanted to have an approach which is formula free. Let metry to explain you.
Noe in the 7 letter word, 2 letters are reptitive. So lets break them in 3+4 letters.
Now, for the first question, ways of selection is either 3C3 X 2+ 3C2 X 3 + 3C1 X 2+ 1= 2+9+6+1= 18 ways.
For the second question, it is selection & arrangement, hence the same shall be written as:3C3 X 2 X 4P4 + (3C2 X 4P4 +3C2 X 2 X 4P4/2!)+ 3C1 X 2 X 4P4/2!+ 1 X 4P4/2!/2!=48+144+72+6=270
I know I have given the nos only but reckon you will understand.
Hi, I have a doubt in Probability...hope it can go here in PnC
Six letters are to be placed in six addressed envelopes. If the letters are placed at random into the envelopes, the probability that
None of the six letters are placed into their corresponding envelopes.
Seven letters are to be placed in seven addressed envelopes. If the letters are placed at random into the envelopes, the probability that
Exactly three of the seven letters are placed into their corresponding envelopes.
In the second one, we need to select 3 letters which goes perfectly into respective envelopes. That can be done in 7C3. Need to know the logic behind remaining 4 going into the wrong ones.
Thanks in advance 😃
we can choose 3 students for the first team in C(9,3) ways.
out of the remaining 6 we can choose 3 in C(6,3) ways.
the remiaining can go into the third team.
so the number of ways of choosing 3 teams of 3 each from 9 students is C(9,3)*C(6,3)=1680.
we can choose a problem out of the three in 3 ways.
so the number of ways in which 3 teams can be formed to study one of the three problems is 1680*3=5040.
in case each team had to study a different problem the answe would have been 1680*3!=10080.
bye..
yew forget to divide by 3! to avoid repetitn of cases ..1680/3! will be the no. of groups..now again mult by 3!
hence the final ans shud be 1680..