Permutations and Combinations

[GauravShah]

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.

[meena8204]

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

[ritika_j]

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....

[MrAnderson]

Originally Posted by ritika_j

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..

[magis_xl]

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 ....

[GauravShah]

Originally Posted by MrAnderson

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.

[GauravShah]

Originally Posted by magis_xl

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

[MrAnderson]

Originally Posted by GauravShah

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.

[magis_xl]

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....

[ritika_j]

Originally Posted by GauravShah

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

Gaurav.

hey the answer is 216... u need to consider 0 as it cant appaer on 1st and 4th postion.