P.S-Sorry for the spam,could not resist it.
IS THIS FOR REAL?? Doc Solving QA!!!!!:-o OLD HABITS DIE HARD EH.
P.S-Sorry for the spam,could not resist it.
P.S-Sorry for the spam,could not resist it.
.Can someone help me out with the following problems with some clear and simple approcah to these qustions
4 balls are to be put in 5 boxes.in how many ways can this be done if
a)Balls are similar and boxes are different
b)balls are different and boxes are similar
c)both boxes and balls are similar
in which of the above cases the formulae (n+r-1)Cr-1 be applied?
is the formulae (n+r-1)Cr or (n+r_1)Cr-1
and which of these formuale is the number of positive integral soultion for
x1+x2+x3+.........xr=n
thanls in advance
a) this is case of putting n identical object in r distinct group
So,no of way=n+r-1Cr-1=8c4=70(as there can be boxes which can be empty)
If, no boxes are empty no of ways are n-1Cr-1, but as there are just 4 balls and 5 boxes, this can never be possible
c)40000,31000,21100,11110,22000->5 WAYS
b)4 balls can be selected in 1 ways->1
3 balls can be selected in 4c3 ways->4
2 balls can be slected in 4c2 ways->6
2 balls can be slected in 4c2 ways->6
1 balls can be selected in 1 ways->1
total ways=18 ways
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
Team a can be selected in 9c3 ways=84 ways
team b can be selected in 6c3 ways=20 ways
Team c can be selected in 1 way
Total way of selection=84*20=1680 ways
Report can be selected in 3c1 ways=3 ways
Total no of ways-1680*3=5040
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...
total no of song=14,
Total no of way 2^12*5c1*6c1=2^12*30
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?
This is a case of selection of n items from r distinct groups
So, 9c3=84
This is a case of selection of n items from r distinct groups
So, 9c3=84
as there is no restriction given on selection there are 4 ways of selecting each disk. hence i think answer should be 4^6.
please correct me if i'm wrong.
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....
Odd=odd+even,
So while selecting first three, two should be odd and 1 should be even
So while selecting next two , 1 should be odd and 1 should be even.
Total no of odd digit=1,5,7=3
Total no of even digit=0,4,8-3
Selcetion of first 3 , 3c2*2c1=6*13=18,
Arranging 3 digit=18*3!=108
But this contain 108/6=18 number where 0 lies at unit place,
So no of ways of slectin first 3 digit=108-18=90
Selcetion of next2=1C1*2c1=2
Arranging 2 digit=2*2!=4
TotAL no of ways=90*4=360
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 ....
1)(a) C(10,1)+C(10,2)+......+C(10,10)
2) i)1st apple-8 children
2nd apple=7 children
3rd apple=6 children
4th apple=5 children
5th apple=4 children
Total ways=8*7*6*5*4
ii)(b) Other 8^5
3)other,3*5!*5!
4)(a) 13^4
5)(a) 13^2
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....
6)(a) 32.32
b)no of ways of selecting black box=32
no of ways of selecting white box=24
total ways=32*24=768
b)other
i)Two digit number not containing 7=4*9=36
Three digit number not containing 7=4*9*9=324=324
Total digit not containing 7=36+324=360
Ttoal no of digit=500-50+1=450+1=451
Total no of digit not containing 7=451-361=90
Others
ii)Total digit=400
number not containing 7=4*9*9=324=324
a324
iii)for 2 digit,
first digit5-4 number
first digit6-3 number
first digit7-2number
first digit8-1 number
for 3 digit,
first digit-1,number(7+6+5+4+3+2+1)=28 number
first digit-2,number=21 number
First digit-3,number=15 number
first digit-4,number=10 number
total number=10+28+21+15+10=84
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 repetition of digits allowed, how many numbers divisible by 4 can be formed when n=4?
For number to be divisible by 4, last two digit should be divisible by 4 or 00
case 1:
So, combinations of last two digit are-40,80,48,84,44,88,04,08
40-5*6=30numbers
80-30numbers
48-30numbers
84-30numbers
44-30 number
88-30 numbers
08-30 numbers
04-30 numbers
Case2,with two 0 at end-25 numbers
with three 0 at end-5 numbers
total numbers=270
Aliter last two place can be filled in 9 ways and first two can be filled in 5*6 ways
total ways=5*6*9=270
The word 'INCONSISTENCY' is jumbles up in such a manner that there is no continuous formation of 'CONSISTENCY' in the so formed words. In how many ways can this be done?
hey ppl solve this with explanation..
No of ways of arranging INCONSISTENCY with no restriction=13!/2!.3!.2!.2!
If CONSISTENCY is taken together, no of ways=3!=6
Total no of ways =13!/2!.3!.2!.2!-6
Hi puys
Please help me with this :
Question : 8 Similar thing to be distributed among 4 different students ?
Answer : 11! / 8! 3!.
======================================
Now when the question iis :
Question : 8 Different things to be distributed among 4 different students?
Answer : 11! / 3! OR 4^8 ?????????
================================
Thanks
Deepak
More questions from P&C;
1) How many 9 letter palindromes are possible from English Alphabets?
2) How many integers are there between 1 and 10^4 that contain exactly one 8 and one 9?
Cheers
1)123456789
positions from 1 to 5 can be taken in 26 ways each,
So no of pallindrome=26^5
If repetition is not allowed,
No of pallindrome=26*25*24*23*22
2)1 nd 10000,abcd
8 and 9 can be arranged in 4 position in 4p2 ways=12 ways
other can be filled in 8*8 ways=64 ways
total ways=64*12=768
ritika_j SaysQ - Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?
7c4-5c2=35-10=25
Hi all,
Could any one of you please give me the solution for the question below.This is eating my brain.
There are 5 letters and 5 distinct addressed envelopes. In how many ways can these letters go into wrongly addressed envelopes.
Kindly post the procedure also.
It is a case of dearrangement,
so, number of dearrangement=n!
=120*(1-1+1/2-1/6+1/24-1/120)
=60-20+5-1=65-21=44 ways
Hi All,
This is my first post in this thread i am having some confusion on one of the questions which i am mentioning below...
Please mention the different approaches that can be taken to arrive to the solution of this problem...
Q:In how many ways can 10 identical presents be distributed among six children so that each child gets atleast 1 present?
It is a case of distributing n identical presents among r groups,
so, no of ways=9C5(n-1Cr-1)
If it would have contained empty group too,
no of ways=16c5(n+r-1Cr-1)
Hi Stalwart,
Thanks, i guess its a problem given in Arun Sharma LOD1 excercise of P&C....;
I also have a confusion regarding this types of questions... Thers this question which says....
A man has 6 friends to send a letter each and he has 4 servants , in how many ways can he distribute his servants so that each of his friends get the letter thru one servant.....
Can u give some concept advice in these kind of questions
this is a case of dividing 6 letter among 4 servant such that no one is empty,
no of ways=5c3 ways=10 ways
aliter
abcdaa->4 ways
abcdab->6 ways
total ways=10
Arun Sharma, LOD I - Pbm 50
Several teams take part in competition, each of which must play one game with all other teams. How many teams took part in competition if they played 45 games in all?
a.5 b.10 c.15 d.20
10,go thru di logic..
Arun Sharma, LOD I - Pbm 50
Several teams take part in competition, each of which must play one game with all other teams. How many teams took part in competition if they played 45 games in all?
a.5 b.10 c.15 d.20
invAMARISH Says10,go thru di logic..
Buddy you solved a question that was posted in 2007!!!!:w00t:
murali13_84 Sayshi soneone pls help me out with these with explanation 1. there are 11 symetric letters and 15 asymetric letters how many 3 letters passwords w/o repetition can be forned with atleast 1 symetric letter
26p3*15p3=12870