Hi Gaurav, I got the same answers. Lets wait for someone to differ. Here are some more....

6) In how many ways can we choose a black and a white square on a chess board?

(a) 32.32
(b) Other

7) In how many ways can we choose a black and a white square on a chess board if the two squares do not belong to the same row or column?

(a) 8.25
(b) Other

8 ) Of the integers 50 to 500 (inclusive) how many integers
(i) contain the digit 7

(a) 91
(b) Other

(ii) are greater than 100 and do not contain the digit 7

(a) 324
(b) Other

(iii) have their digits in strictly increasing order.

(a) 124
(b) Other

More coming....

I got the answer as 360...haven't verified if 2*(9*4p3-6*3p2) = 360
Gaurav.

It is..i was slightly confused abt the formula for permutations
i guess 4p3 is 4*3*2 ...24 and 3p2 is 3*2 ..
tht wud make it 2*(9*24 - 6*6)= 180 *2 = 360....
wht abt the aproach gaurav...did u do it the same way or do u hv another way of doing it?
andy.

Hi All

I am a new entrant to PG. I have gone through the posts. Here are some intresting PC questions. Please give your answers and working for common benefit. Thank You.

1) How many ways can a committee of k people be chosen from 10 people if k can be 1,2,3...or 10.

(a) C(10,1)+C(10,2)+......+C(10,10)
(b) 10+10.9+10.9.8+ ..... + 10.9.8....3.2.1
(c) Other

2) A mother distributes 5 different apples among 8 children.
(i) How many ways can this be done if each child recieves at most one apple?

(a) 8.7.6 (5!)
(b) Other

(ii) How many ways can this be done if there is no restriction on the number of apples a child can recieve?

(a) 5^8
(b) Other

3) A compartment has 10 seats - 5 seats facing the engine and the other 5 facing away from the engine. Of the 10 passengers, 4 prefer to face the engine, 3 prefer away and 3 have no preference. In how many ways can the passengers be arranged on their preference?

(a) 3
(b) Other

4) In how many ways can 4 cards of different suits be selected from a deck of cards?

(a) 13^4
(b) Other

5) In (4) in how many ways can the 4 caqrds be selected where diamond selection has same value as the clubs selection and spade that of heart?

(a) 13^2
(b) Other

More to follow ....

Here are my answers

1) (a) C(10,1) + ...

2) (i) (a) 8.6.7.(5!)
(ii) Not solved yet

3) (b) Other. Answer wud be 3C1 * 5! * 5!

4) (a) 13^4

5) (a) 13^2

Hope these are correct

Cheers

Gaurav

A=={0,1,4,5,7,8}
Unless stated otherwise number are formed without repetition of digits. In the following question'n' refers to the number of digits in the number being formed.

Q- with n=5 how many numbers can be formed such that the sum of three-digit number formed withe first three digits and the two-digit numbers formed with the remaining two digits is odd?

hey ppl. slove with explanation....

hi ..interesting problem..
this is my solution ..the numbers might be slightly wrong but i think the approach is right..
here goes...

basically u hv 6 numbers to be placed in 5 slots ..satisfying ur condition ..
for satisfying the condition..i.e,sum of first 3 +last 2 digits is odd...there are two ways..
number formed by first 3 is odd and next 2 is even
or vice versa...
for first case...
spose ur no is abcde
therefor c can be filled in 3 ways(1,5,7) e can be filled in 3 ways (0,4,8 )
and the remaing 3 spots can be filled by the remaining 4 nos in 4p3 ways
so ... total no of ways = 3*3*4p3 ..
now we have to subtract becos there will be nos with 0 in the first place..
for that we can proceed like this ..place 0 at first position..
as above 3rd spot can be filled in 3 ways ..but last sot can e filled in only 2 ways (4,8 ) ..now remaining 2 spots ad 3 nos ..can be done in 3p2 ways..
so total....3*2*3P2

for second case using the same logic as above..
3*3*4p3 - 2*3*3p2 ..

so total number of nos ... 2*(9*4p3-6*3p2)

..any flaw in the logic??do point out.
andy..

I got the answer as 360...haven't verified if 2*(9*4p3-6*3p2) = 360

Gaurav.

Hi All

I am a new entrant to PG. I have gone through the posts. Here are some intresting PC questions. Please give your answers and working for common benefit. Thank You.

1) How many ways can a committee of k people be chosen from 10 people if k can be 1,2,3...or 10.

(a) C(10,1)+C(10,2)+......+C(10,10)
(b) 10+10.9+10.9.8+ ..... + 10.9.8....3.2.1
(c) Other

2) A mother distributes 5 different apples among 8 children.
(i) How many ways can this be done if each child recieves at most one apple?

(a) 8.7.6 (5!)
(b) Other

(ii) How many ways can this be done if there is no restriction on the number of apples a child can recieve?

(a) 5^8
(b) Other

3) A compartment has 10 seats - 5 seats facing the engine and the other 5 facing away from the engine. Of the 10 passengers, 4 prefer to face the engine, 3 prefer away and 3 have no preference. In how many ways can the passengers be arranged on their preference?

(a) 3
(b) Other

4) In how many ways can 4 cards of different suits be selected from a deck of cards?

(a) 13^4
(b) Other

5) In (4) in how many ways can the 4 caqrds be selected where diamond selection has same value as the clubs selection and spade that of heart?

(a) 13^2
(b) Other

More to follow ....

A=={0,1,4,5,7,8}
Unless stated otherwise number are formed without repetition of digits. In the following question'n' refers to the number of digits in the number being formed.

Q- with n=5 how many numbers can be formed such that the sum of three-digit number formed withe first three digits and the two-digit numbers formed with the remaining two digits is odd?

hey ppl. slove with explanation....

hi ..interesting problem..
this is my solution ..the numbers might be slightly wrong but i think the approach is right..
here goes...

basically u hv 6 numbers to be placed in 5 slots ..satisfying ur condition ..
for satisfying the condition..i.e,sum of first 3 +last 2 digits is odd...there are two ways..
number formed by first 3 is odd and next 2 is even
or vice versa...
for first case...
spose ur no is abcde
therefor c can be filled in 3 ways(1,5,7) e can be filled in 3 ways (0,4,8 )
and the remaing 3 spots can be filled by the remaining 4 nos in 4p3 ways
so ... total no of ways = 3*3*4p3 ..
now we have to subtract becos there will be nos with 0 in the first place..
for that we can proceed like this ..place 0 at first position..
as above 3rd spot can be filled in 3 ways ..but last sot can e filled in only 2 ways (4,8 ) ..now remaining 2 spots ad 3 nos ..can be done in 3p2 ways..
so total....3*2*3P2

for second case using the same logic as above..
3*3*4p3 - 2*3*3p2 ..

so total number of nos ... 2*(9*4p3-6*3p2)

..any flaw in the logic??do point out.
andy..

A=={0,1,4,5,7,8}
Unless stated otherwise number are formed without repetition of digits. In the following question'n' refers to the number of digits in the number being formed.

Q- with n=5 how many numbers can be formed such that the sum of three-digit number formed withe first three digits and the two-digit numbers formed with the remaining two digits is odd?

hey ppl. slove with explanation....

Your approach is converted into that formula...it takes time to do this process ..so convert it into formulae results n+r-1 C r-1

The number of ways of partitioning n non-distinct things into r distinct gropus (some grp may be empty) is n+r-1 C r-1...so aplly this in the problem....

Pendayal had before given a diffrent answer but his mistake was he tght all n things are distinct.....but they are not distinct ...
he applied this formula:

The number of ways of dividing n distinct things into r distinct groups(where some may be empty)=r^n

-meena

Just logged in back....

Thanx meena for that elabotation

hey rikita... didn't get it how did the formula work... pls someone clarify

Cheers

Gaurav.

A bin of computer disks contain a supply of disks from 4 different
manufacturers. In how many ways can you choose 6 disks from the bin?

i think this can be done by using this formula (n+r-1)C(r-1)=(6+4-1)c(4-1)
=9C3=84

thanx meena for that clarification.

bye...

Pendyal,
I think you have all the permutations, but what we require here, are the combinations.
I am getting the answer using the following approach:

1) 6 disks are from the same manufacturer(M) = 4 cases
2) Cases with 2 Ms :
a) 5 --> M1 & 1 --> M2 = 12 cases
b) 4 --> M1 & 2 --> M2 = 12 cases
c) 3 --> M1 & 3 --> M2 = 6 cases
3) Cases with 3 Ms:
a) 4 --> M1 & 1 --> M2 & 1 --> M3 = 12 cases
b) 3 --> M1 & 2 --> M2 & 1 --> M3 = 24 cases
c) 2 --> M1 & 2 --> M2 & 2 --> M3 = 4 cases
4) Cases with 4 Ms:
a) 3 --> M1 & 1 --> M2 & 1 --> M3 & 1 --> M4 = 4 cases
b) 2 --> M1 & 2 --> M2 & 1 --> M3 & 1 --> M4 = 6 cases

Hence, adding all the above possible cases,
= 12 + 12 + 6 + 12 + 24 + 4 + 6
= 84

As, the solution was provided, it made my life much easier

If the above solution seems correct, please reply, so that if required, I can elaborate further.

Cheers

Gaurav

i didnt understand the solution.cud u please elaborate.
(more specifically in the third case of 2 manufacturers how u got 6 ways and in the other 2 cases how u arrived at the possible combinations was all greek to me.)

I think its rite..see if my explantaion is ok....

as he says:

1) 6 disks are from the same manufacturer(M) = 4 cases
all are eitehr from M1 or M2 or M3 or M4

M1 M2 M3 M4
6 0 0 0
0 6 0 0
0 0 6 0
0 0 0 6



2) Cases with 2 Ms :

You select two manufaturers and from the two you can get different combinations like (5,1) (4,2) (3,3)

therefore for (5,1)

M1 M2 M3 M4
5 1 0 0
5 0 1 0
and so on...
Formaula is :select 2 out of 4 manufaturers 4C2
now the two can be intercahnged that is (5,1) to (1,5)
therefore multiply by 2
===>4C2*2= 12


same way for (4,2)
M1 M2 M3 M4
4 2 0 0
4 0 2 0
and so on

same way as before
====>4C2*2=12

for (3,3)
select two 4C2....interchanging (3,3) to (3,3) doesn't matter

hence
====>4C2=6


a) 5 --> M1 & 1 --> M2 = 12 cases
b) 4 --> M1 & 2 --> M2 = 12 cases
c) 3 --> M1 & 3 --> M2 = 6 cases

3) Cases with 3 Ms:

a) 4 --> M1 & 1 --> M2 & 1 --> M3 = 12 cases
select 3 = 4C3
(4,1,1)-(1,4,1)-(1,1,4) two are 1,1 so repetitions in 3! so divide by 2!
3!/2! = 3


b) 3 --> M1 & 2 --> M2 & 1 --> M3 = 24 cases
4C3*3! (all are differnt number 3,2,1
====>24
c) 2 --> M1 & 2 --> M2 & 2 --> M3 = 4 cases
select 3 = 4C3
====> 4




4) Cases with 4 Ms:
a) 3 --> M1 & 1 --> M2 & 1 --> M3 & 1 --> M4 = 4 cases
4!/3! =4
====> 4
b) 2 --> M1 & 2 --> M2 & 1 --> M3 & 1 --> M4 = 6 cases
4!/(2!*2!)
====> 6

so add up .....u get the answer

-meena

hey it seems right as the official answer is 84 .... thatz what i had mentioned earlier ....
please elaborate...

Pendyal,
I think you have all the permutations, but what we require here, are the combinations.
I am getting the answer using the following approach:

1) 6 disks are from the same manufacturer(M) = 4 cases
2) Cases with 2 Ms :
a) 5 --> M1 & 1 --> M2 = 12 cases
b) 4 --> M1 & 2 --> M2 = 12 cases
c) 3 --> M1 & 3 --> M2 = 6 cases
3) Cases with 3 Ms:
a) 4 --> M1 & 1 --> M2 & 1 --> M3 = 12 cases
b) 3 --> M1 & 2 --> M2 & 1 --> M3 = 24 cases
c) 2 --> M1 & 2 --> M2 & 2 --> M3 = 4 cases
4) Cases with 4 Ms:
a) 3 --> M1 & 1 --> M2 & 1 --> M3 & 1 --> M4 = 4 cases
b) 2 --> M1 & 2 --> M2 & 1 --> M3 & 1 --> M4 = 6 cases

Hence, adding all the above possible cases,
= 12 + 12 + 6 + 12 + 24 + 4 + 6
= 84

As, the solution was provided, it made my life much easier

If the above solution seems correct, please reply, so that if required, I can elaborate further.

Cheers

Gaurav

i didnt understand the solution.cud u please elaborate.
(more specifically in the third case of 2 manufacturers how u got 6 ways and in the other 2 cases how u arrived at the possible combinations was all greek to me.)

A bin of computer disks contain a supply of disks from 4 different
manufacturers. In how many ways can you choose 6 disks from the bin?

hey solve with explanation........

each disk chosen from the bin can be chosen in 4 ways.

so the 6 disks can be chosenin 4*4*4*4*4*4 ways i.e 4^6 ways.

bye..

Pendyal,
I think you have all the permutations, but what we require here, are the combinations.
I am getting the answer using the following approach:

1) 6 disks are from the same manufacturer(M) = 4 cases
2) Cases with 2 Ms :
a) 5 --> M1 & 1 --> M2 = 12 cases
b) 4 --> M1 & 2 --> M2 = 12 cases
c) 3 --> M1 & 3 --> M2 = 6 cases
3) Cases with 3 Ms:
a) 4 --> M1 & 1 --> M2 & 1 --> M3 = 12 cases
b) 3 --> M1 & 2 --> M2 & 1 --> M3 = 24 cases
c) 2 --> M1 & 2 --> M2 & 2 --> M3 = 4 cases
4) Cases with 4 Ms:
a) 3 --> M1 & 1 --> M2 & 1 --> M3 & 1 --> M4 = 4 cases
b) 2 --> M1 & 2 --> M2 & 1 --> M3 & 1 --> M4 = 6 cases

Hence, adding all the above possible cases,
= 12 + 12 + 6 + 12 + 24 + 4 + 6
= 84

As, the solution was provided, it made my life much easier

If the above solution seems correct, please reply, so that if required, I can elaborate further.

Cheers

Gaurav

hey but OA is 84

A bin of computer disks contain a supply of disks from 4 different
manufacturers. In how many ways can you choose 6 disks from the bin?

hey solve with explanation........

each disk chosen from the bin can be chosen in 4 ways.

so the 6 disks can be chosenin 4*4*4*4*4*4 ways i.e 4^6 ways.

bye..

A bin of computer disks contain a supply of disks from 4 different
manufacturers. In how many ways can you choose 6 disks from the bin?

hey solve with explanation........

There are 5 Rock songs, 6 Carnatic songs and 3 Indi pop songs. How many different albums can be formed using the above repertoire if the albums should contain at least 1 Rock song and 1 Carnatic song?


Solve with proper explanation...

first choose a rock song and a carnatic song.this can be done in 5 ways and 6 ways respectively.
so we can choose a rock song and a carnatic song together in 30 ways.

now of the remaining 12 songs,each song can either be in the album or not be in the album.that is there are 2 ways for each of the 12 songs.
so these 12 songs can be chosen in 2*2*2....*2 (i.e. 2^12).

the total no. of ways becomes 30*(2^12).


bye..

There are 5 Rock songs, 6 Carnatic songs and 3 Indi pop songs. How many different albums can be formed using the above repertoire if the albums should contain at least 1 Rock song and 1 Carnatic song?


Solve with proper explanation...

1680*3 was only a way to come up with one of the options as the answer.


if i were not given the options i wud have given 1680*3! as the answer.

Hey pendyal,

Don't you think a more appropriate answer is the latter one you gave i.e. 1680*3! ( though it is not in the options)?

Also, i think we can even stop at 1680 because the question says "How many teams can be formed?".

But 1680*3 doesn't seem to be too convincing.

Nine students are split into three equal teams to develop reports on one of three problems:

shortage of skilled labor, violence in schools, and low standardized test scores. How many

different teams of students are possible?

(A) 5040

(B) 1680

(C) 1512

(D) 504

(E) 168

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..

Nine students are split into three equal teams to develop reports on one of three problems:

shortage of skilled labor, violence in schools, and low standardized test scores. How many

different teams of students are possible?

(A) 5040

(B) 1680

(C) 1512

(D) 504

(E) 168

hi vinayak
is the formulae for the following
The number of positive integral solution(natural number) for
x1+x2+x3+.........xr=n is (n-1) C (r-1) or (n-r+1)C(r-1)

Hi
I seem to have a little confusion between a) and c).I feel they are the same .In other words, in a) all that is important is to choose 4 boxes from 5.Why is it important , the order in which you put the balls in the boxes as the balls are all similar ?i feel there is no permutation involved....I know this is the wrong answer but im not able to convince myself...so can somebody help....?

a)
as all 4 balls are same so if we arrnage them among them selves then onli one way , let me explain it further
lets take 4 rs1 coins to b given to 4 kids n each kid shd b given 1 rs exactly , there is only one way u can give , it doesnt matter which coin u give to which boy , end result will b same as all coins are identical .

we can arrange these 4 IDENTICAL balls in 5 DIFF boxs in 5x4x3x2
u can derive this very simply lik this : 1st red ball will hv 5 boxes to choose , or in other words we can put 1st red ball in any of 5 diff boxes, so wen1st red ball takes 1 box then 2nd red ball can b put in ny of remaining 4 diff bxes , simlilarly 3 rd ball can b put in ny of remaining 3 boxes , 4th ball in ny of remaining 2 boxes
so we get 5x4x3x2
we put x( multiplication sign ) in between 5 4 3 n 2 bc we r using word AND
we will read e solution lik : we can put 1st red ball in any of 5 diff boxes AND 2nd red ball can b put in ny of remaining 4 diff bxes AND 3 rd ball can b put in ny of remaining 3 boxes AND 4th ball in ny of remaining 2 boxes
so ans is 5x4x3x2x1( this 1 is for 1 way of arranging balls )

b)
4 diff balls n 5 same boxes , so in this case boxes wont matter , balls will
we can choose 4 identical boxes out of 5 identical boxes in 5c4 ( if there were only 4 identical boxes n we hv to choose 4 out of them then we whd hv had only one way of doing it , we can prove it by formula also 4c4 = 1 .....) tat is equal to 5 AND we can arrange 4 DIFF balls in 4! ways
as we used AND in above line , e solution wil b like
5 x 4!

c)
all balls same , so ball can b arranged in 1 way only n as all boxes same so we can choose 4 identical boxes out of 5 identical boxes in 5c4=5
so ans is
1x5=5

i dont noe abt formula u mentioned above , our faculty told us to take step by step approach or else this topic can b highly confusing