6 different types of fruit .
Each type have 5 fruits .
No . of ways to select 1 or more fruits ?
Question :
6 different types of fruit .
Each type have 5 fruits .
No . of ways to select 1 or more fruits ?
6 different types of fruit .
Each type have 5 fruits .
No . of ways to select 1 or more fruits ?
how come each wrestler plays (m-3) matches, each would play (m-2) matches.. isnt it?
pls explain..
Question :
6 different types of fruit .
Each type have 5 fruits .
No . of ways to select 1 or more fruits ?
It's 6^6-1.
per fruit, essentially, there are 6 choices, as we can select 0, 1, 2...5 fruits of one type. For six fruits, there are 6^6 choices, one of which is the case in which we've not chosen any fruit of any type. So subtracting that case, we're left with 6^6-1 possible ways.
@ chillfactor
Sir ..
In the second question below ..
why 1 0 6 case is not included along with ( 1 1 5 , 1 4 2 , 1 3 3 ) ????
2) 3*2^7 - 3 = 384 - 3 = 381
3*2^7 are the cases when atleast one of them haven't got anything, but few cases are repeated since here all those cases are also there when two of them haven't got anything
3 are the cases when only one of them got all objects
_______________________________________________________________
3) Three cases:-
i) (1, 1, 5) - 3*C(7, 1)*C(6, 1) = 126 ways
3 because 5 can be given to any one of the three
ii) (1, 2, 4) - 6*C(7, 1)*C(6, 2) = 630
6 because (1, 2, 4) can be permuted in 3! ways for three persons
iii) (1, 3, 3) - 3*C(7, 1)*C(6, 3) = 420
3 because 1 can be given to any one of the three
Total = 126 + 630 + 420 = 1176
@ chillfactor
Sir ..
In the second question below ..
why 1 0 6 case is not included along with ( 1 1 5 , 1 4 2 , 1 3 3 ) ????
Yup, you are correct. It should be there along with (0, 2, 5), (0, 3, 4). Somehow I assumed that everyone should get at least one
help on this
wat is the probability of getting a word with all 'T's together for the word ATTEMPT?
i am getting 1/42...... wrong???
help on this
wat is the probability of getting a word with all 'T's together for the word ATTEMPT?
i am getting 1/42...... wrong???
What is the answer given as?? I got 1/42 in the 1st attempt too.:D
spectramind07 SaysWhat is the answer given as?? I got 1/42 in the 1st attempt too.:D
spectra bhai, i dont know where did my friend found this problem. He told the given answer is 1/7
help on this
wat is the probability of getting a word with all 'T's together for the word ATTEMPT?
i am getting 1/42...... wrong???
Total no. of ways in which in it could be arranged = 7!/3!
When all 'T's are together, no. of ways = 5!
So, 5!*3! / 7! = 1/7.
Total no. of ways in which in it could be arranged = 7!/3!
When all 'T's are together, no. of ways = 5!
So, 5!*3! / 7! = 1/7.
Hahahah,so right
help on this
wat is the probability of getting a word with all 'T's together for the word ATTEMPT?
i am getting 1/42...... wrong???
It should be 1/7(and i got it right in the first attempt!
)there are 5*4! ways in which the three T's are together.
Total combinations possible are 7!/3!
So, probability is 1/7.
Students score in each section is a, b, c, d, then
a + b + c + d = 90
=> C(93, 90) ways, but we have counted those cases also when a, b, c, d > 45
When a > 45, i.e., a = 46 + a'
=> a' + b + c + d = 44
=> C(47, 44) ways
Similarly for other variables
So, total number of ways = C(93, 90) - 4*C(47, 44) = 64096
i have a doubt....
the equation used is a+b+c+d=90
dont you think this is wrong
here we are only finding the no of cases in which the student gets 90
however, we have to find the no of cases when
90
the question is....
The AMS MOCK CAT test CATALYST 19 consists of 4 sections. Each section has a maximum of 45 marks. Find the number of ways in which a student can qualify in the AMS MOCK CAT if the qualifying marks is 90.
1) 36546
2) 6296
3) 64906
4) n.o.t
Total no. of ways in which in it could be arranged = 7!/3!
When all 'T's are together, no. of ways = 5!
So, 5!*3! / 7! = 1/7.
Shit man!!!!!!!!!!!!!! such a silly mistake.. agar exam me aata hota do -4 for me and spectra :banghead:
thanku messy
the question is....
The AMS MOCK CAT test CATALYST 19 consists of 4 sections. Each section has a maximum of 45 marks. Find the number of ways in which a student can qualify in the AMS MOCK CAT if the qualifying marks is 90.
1) 36546
2) 6296
3) 64906
4) n.o.t
getting it as 45!. NOT
:banghead:
the question is....
The AMS MOCK CAT test CATALYST 19 consists of 4 sections. Each section has a maximum of 45 marks. Find the number of ways in which a student can qualify in the AMS MOCK CAT if the qualifying marks is 90.
1) 36546
2) 6296
3) 64906
4) n.o.t
is it 64906?
chandrakant.k Saysgetting it as 45!. NOT
jain4444 Saysis it 64906?
the answer as discussed before is 64906....but someone please clear my doubt.....
Problem Set from Arun Sharma
From 4 men and 4 ladies, a committee of 5 is to be formed.
a) find no. of ways of doing so if comity conststs of a presi, a vice presi and three secs.
b) what will b the number of ways of selecting the committee wid at least 3 women such that at least one woman holds the post of either presi or vp.
c) find the no. of ways of selecting the committee with a max of 2 women and having at the max one woman holding one of the two posts on the committee
From 4 men and 4 ladies, a committee of 5 is to be formed.
a) find no. of ways of doing so if comity conststs of a presi, a vice presi and three secs.
b) what will b the number of ways of selecting the committee wid at least 3 women such that at least one woman holds the post of either presi or vp.
c) find the no. of ways of selecting the committee with a max of 2 women and having at the max one woman holding one of the two posts on the committee
Problem Set from Arun Sharma
From 4 men and 4 ladies, a committee of 5 is to be formed.
a) find no. of ways of doing so if comity conststs of a presi, a vice presi and three secs.
b) what will b the number of ways of selecting the committee wid at least 3 women such that at least one woman holds the post of either presi or vp.
c) find the no. of ways of selecting the committee with a max of 2 women and having at the max one woman holding one of the two posts on the committee
a)8*7*5c3=560
b)34??
c)36???
Problem Set from Arun Sharma
From 4 men and 4 ladies, a committee of 5 is to be formed.
a) find no. of ways of doing so if comity conststs of a presi, a vice presi and three secs.
b) what will b the number of ways of selecting the committee wid at least 3 women such that at least one woman holds the post of either presi or vp.
c) find the no. of ways of selecting the committee with a max of 2 women and having at the max one woman holding one of the two posts on the committee
hey man why are posting again this here,
you didnt get the solution at quant thread?????