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.
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!
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
@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...
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!
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 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
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
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..