Permutations,combinations Plus Probability...Some Awesome questions

[The Villain]

Q1..8 different books and 2 identical copies of each.Find no of ways in which 1 or more books can be selected?
Q2..If 35 fruits are distributed among 3 patients,then probability no one gets less than 10 is
a)5/111
b)7/111
c)5/222
D)7/222

q3..A and B throw a die each.The probability that A's throw is greater than B's is
a)7/12
b)5/12
c)1/6
d)1/2

q4...There are two teams with n persons in each.The probability of selecting 2 persons fr 1 team and 1 person from other team is 6/7
then n is
a)3
B)4
c)5
d)6

q5....A nd B stand in a ring with 10 other persons .the prob hat exactly 3 persons are between A and b is
a)3/11
b)2/11
c)3/22
d)none of the above

q6...What is the possibilty of placing n-1 letters correctly in envelope?
a)1/n
b)1/n-1
c)n/nfactorial
d)none of the above


q7...there are 5 boys and 5 girls...no of ways in which boy and girl can sit alternately?
a)5 factorial
b)5 factorial*6P5
c)5 fact(5fact+5fact)
d)none of the above

[Arushi :))]

I have been able to do these.
3-b
4-b
5-b
7-c

[jack]

for Q-6
Exhaustive no. of case=(n-1)!
favorable no. of case =
If we choose 1 letter, then it goes 1 correct envelop
so its probability is 1
then remaining (n-2) envelopes can go in (n-2)!
so,
P(a letter goes into the right envelop)=(n-2)!/(n-1)!
1/n-1
option b
But i am not 100% sure

[jack]

favourable no. of case (2,1),(3,1),(3,2),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(5,4),(6,1),(6,2),(6,3),(6,4),(6,5)
required probability=15/36
5/12

[SINGHAM]

Hope your questions contained more details! But, thanks for so good questions!

1. I think we have to select only one copy of any book. So, we have to select at least one book and we can select any book in 2 ways as there are 2 copies. So, we have (8C1)*2+...+(8C8)*(2^8)=(2+1)^8-1.

2.Numerator: we can't distinguish between fruits. thus, first give 10 fruits each. then, we have to distribute 5 fruits among 3 or equivalently, 8 fruits among 3 so that each gets at least 1. thus, 7C2
Denominator: we have to distribute 35 fruits among 3 or equivalently, 38 fruits among 3 so that each gets at least 1. thus, 37C2
Doing algebra, we have 7/222

3. Numerator: 6X6 square matrix, delete the diagonal of 6 elements and divide by 2: we get 15.
Denominator:6*6
Sol:15/36=5/12

4.it is easy (4C2*4C1)/8C3

5.ITS RING:
Numerator: Fix relative positions of A & B and count them as single person. So, we have to arrange 11 persons but in ring. So (11-1)! ways. But A and B can be arranged 2! ways mutually. Thus 2!*10!
Denominator: (12-1)!=11!
Thus, 2/11

6.I think the correct question should be:" What is the possibility of placing exactly n-1 letters correctly in n envelopes?"

And this is tricky question! Nothing needed other than little thought: If we are to place correctly n-1 letters in n envelopes, so we have to place that last letter in the left envelope, but as all letters were placed correctly, so the last envelope is also correct for that letter.

So,the possibility of placing exactly n-1 letters correctly in n envelopes is 0

7. First let the boys sit, in 5! ways. then _B_B_B_B_B_ 6 places are for girls. Girls can choose 6C5 places, and can arranged in 5! ways so the answer is 5!*6C5*5!=5!*6P5

[The Villain]

ans
q1 3^8-1
q2 b
q3 b
q4 b
q5 b
q6 d
q7 c

[SoniaKapoor]

Pls help me with quest 1...

[SoniaKapoor]

Ron. . .Do you have more quest on these..it will really be helpful if you could post them

[The Villain]

@I have quite a lot.But its not possible to post all of them..

[SINGHAM]

How is answer of 7 is C ? Please help!

[SINGHAM]

Ok, got it!

In order to actually sit alternatively, there are only two such ways; either GB GB GB GB GB or BG BG BG BG BG. Thus, it is 2(5!)^2=5!(5!+5!)

[gayathri]

can you explain q3 ?? pls

[SINGHAM]

It is very simple.

Think A throws first and B next. So sample space is  of 6X6 elements. Out of which, you have to take only those elements in which first one is greater than second one. You can solve this by listing all of them, as Jack did, or simply go this way: " 6X6 square matrix, delete the diagonal of 6 elements and divide by 2: we get 15." and you get answer!

[phelps.phan]

Hi Ron/Singham,
can u tell how to solve the below one

In how many ways can 10 identical presents be distributed among 6 children so that each child gets at least one present.
a) 15C5 b)16C6 c)9C5 d)6^10

[The Villain]

@ Phelps
the no of ways is same as number of natural no of solns of eqn
a+ b+c+d+e+f=10
which is 9C5

[phelps.phan]

Thanks Ron, the answer is right, but I'm unaware of the natural no of solns of a eqn. Which chapter/topic of any books is it explained? Or can u tell how the number of solns for that eqn is 9C5. Thanks :)

[The Villain]

@Phelps
a+b+c+d+e+f=10
so..here n is 10 and r is 6 (count a, b, c , d ,e, f)
so no of solns is n-1 C r-1 =9C5
Hpe you hot it man:-)

[phelps.phan]

thanks Ron, got it

[RajEco]

The best one was Q6. :P

There is no way in which n-1 letters can be placed correctly. putting n-1 letters correctly implies that only 1 letter is being placed wrongly, which cannot happen because if one letter is put in some other place it should mean that there is another letter which is not in the right place.

[RajEco]

Can any ecplain q2 again.

@Singham I saw ur solution but its not clear to me. Can you elaborate please. How you used 7C2 and 37C2