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 similar to pbm no 4
visualize this as - distribute n things among r things --> (r)n n=6 and r=4
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?
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
1 + 2+ ..... ( n ) = 45 ==>> n = 9
==>> Number of players is 10 ( One cant play with himself .:wow:. )
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?
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...
Is the sequence how they appear in any album is immaterial ... then number of all possible such albums are ..... ( Keeping in mind that each song in a particular catagory is a different one)
hi 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
hi 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
hey i am not very sure about the soln but i guess some thing on the following lines: total no of letters = 26 3 letter pswd wo rptn with 26 letters = 26P3 3 letter pswd wo symmetric letter = 15P3 thus 3 letter pwsd with atleast one symmetric letter = (26P3 - 15P3)
How many numbers between 200 and 1200 can be formed with digits 0,1,2,3 (no repetition of digits allowed)
a. 6 b. 8 c. 2 d. 14 e. 16
I am able to come with only 6 numbers -- 1023,1032,0231,0213,0312,0321
Correct ans is 14 ... which are other nos and what is the technique
this is a good one i think the soln goes as follows: here you are implicitly assuming that the nos have to be 4 digit with permission of using 0 in the units place. however there is no such restriction thus we have
3 digits no formed with these 4 nos : 2*3P2 = 12 (2 is multiplied because the first digit can either be 2 or 3)
4 digits no formed with these 4 nos are : 1*1*2! = 2 ( since 3 digits already cater for the nos formed with 0 at the thousands place) this total is 12 + 2 = 14:grin:
Puys I hv a doubt in the following Qn In how many ways can a person distribute greeting cards (obviously not identical ones) to 6 different friends when he has 4 servants to do so. The answer is 4^6 Explaination - every cards has 4 options to chose from so 4 x 4 x 4 x 4 x 4 x 4= 4^6 My doubt is : why cant we go the other way around namely every servant has 6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = 2^6 options now there r 4 servants so 4 x 2^6 = 4^4 which is incorrect. Can someone plz point out the error. I'm eagerly waiting for someone to clarify it. Thank u.
can someone plz help me with this one ? Out of 4 men and 4 women a committee of 5 pple has to be created comprising a president a vice president and 3 secretaries. Find the number of ways of forming this committee such that there are at least 3 women in the committee and at least one woman holds the post of the president or the vice president. Puys the answer given is (4C4 x 4C1 x 5!/3!) +(4C2 x 4C3 x 5!/3!) - (4C2 x 4C3 x 2C2 x 2!) puys I understood what and why they hv done in the 1st & 2nd brackets but I don't understand wat hv they done in the 3rd bracket. Someone plz help !!
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
let there be x teams , total number of matches will be (x-1)+(x-2)+(x-3)-----+1 so x(x-1)/2=45 so x comes out to be 10 (b)
hey it seems right as the official answer is 84 .... thatz what i had mentioned earlier .... please elaborate...
correct answer is 84 .... the disks may be chosen either all from same manufrer or all diff manfs .. and so on so u need to diffenciate in these cases and calcuklatye cases separately .
the 4^ 6 way wount work coz u r considering xtra cases in that approach .... 1) all diff mans 4 ways (4C1) 2)from 2 mans .. 2 mans in 4C2 ways X5 to arrange em (suppose a n b) (AB two) then 4 remain (AAAA,BBBB,AABB,AAAB,BBBA) 3)4C3 X10 ways (10 similarly as above...) 4) (4C2+4) total 4 +30+40+10 84 ways
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
let A B C D be 4 manufacturers .. 6 disks may be either from 1 2 3 or all 4 manufcs..
1. single manufacturer . in 4 ways 2 . two manufas.. in 4C2X 6 ways 3. from 3 manfs.. in 4C3X10 4. all diff 1X10
adding total 90 ways..
This is incorrect... The problem is equivalent to saying out of six chosen CD' s in how many ways can you categorize them into groups of 4 i.e., 6=a+b+c+d where a is no of brand A and B is no of brand B items and so on .... Categorising 6 into 4 groups can be done as follows _ _ _ _ _ _ _ _ _ In the above blanks if you put three separators ( |'s) in any of the 9 blanks you will be categorizing 6 items into 4 groups So 9C3 i.e., choose 3 positions out of 9 for separators as all the separators are same and no arrangement of separator's needs to be considered.
ya actually i calculated the answer wrong in the previous reply but i corrected the answer to 84 ... actually both ways to appraoch the problem r correct but (n+r-1)C(r-1) approach is cerainly better as it is short but other one is a more visulaized answer .. apologies for the pevious incorrect post ..
correct answer is 84 .... the disks may be chosen either all from same manufrer or all diff manfs .. and so on so u need to diffenciate in these cases and calcuklatye cases separately .
the 4^ 6 way wount work coz u r considering xtra cases in that approach .... 1) all diff mans 4 ways (4C1) 2)from 2 mans .. 2 mans in 4C2 ways X5 to arrange em (suppose a n b) (AB two) then 4 remain (AAAA,BBBB,AABB,AAAB,BBBA) 3)4C3 X10 ways (10 similarly as above...) 4) (4C2+4) total 4 +30+40+10 84 ways
There are 6 sticks of different heights which are to be arranged in two rows of 3 each such that the sticks are in ascending order from left to right in each row, and the stick in front is always smaller than the one at the back. How many such arrangements are possible? guys the answr is... 5 but i dont know the method to do it except making all such arangements... is there any short way?
ps: I've also posted the same post in the other P&C; forum coz i dint know which one is active
There are 6 sticks of different heights which are to be arranged in two rows of 3 each such that the sticks are in ascending order from left to right in each row, and the stick in front is always smaller than the one at the back. How many such arrangements are possible? guys the answr is... 5 but i dont know the method to do it except making all such arangements... is there any short way? ps: I've also posted the same post in the other P&C; forum coz i dint know which one is active
Hi, I have a question on permutation and combination.
Consider 6 red, 5 blue and 3 yellow pair of socks in a drawer. Electricity is not available in that room. How many time you need to take to get a pair of socks from each color?