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
the answer is 8*7*6*5*4 right?
please correct me if iam wrong
@rkshtsurana said: for the first one it is coming 21(10-1)! / 4!*6!but for the next part--a fraction
first fix one R or Green ball and permutate the other 9 in 9! ways.
as Red and green balls are similiar so total ways
9!/4!*6!
but the ball we fixed in begining can be choosen in 2 ways so
ans =2*21=42
as Red and green balls are similiar so total ways
9!/4!*6!
but the ball we fixed in begining can be choosen in 2 ways so
ans =2*21=42
hey puys,
can anyone pls suggest a good book to learn pnc basics from?? should i try school level books??
Puys,Does anybody have a link to illustrate how to get the rank of a word? Or can anyone explain with an example?
@[576959:Stoicalme] 1) What do you mean by Finding out a rank of the word
Ans : It is finding out the position of the word when all possible words have been formed using all the letters(only once) in alphabetical order .
Ex : let us find the rank of word POINT( from time material )
Now arrange these in alphabetical order
I N O P T
Count all the words that starts with I i.e 4!
now similarly N 4! and O 4!
Now come to P
P****
now we have P the second letter again arrange it in alphabetical order.
So for I 3! = 6
for N 3! = 6
after N we have O which is the case for our word. So go for 3rd letter arrange in alphabetical i.e O which is same as our word. next N next T. So we have arrived at our word POINT
Now add up all the combinations above
i.e 4!+4!+3! = 24+24+24+6+6 = 84
so rank of a word is 85
Hope you understand
Badly needed a thread dedicated to PnC.
Thanks. Will keep tracking this.
Can anyone please help me in the detailed analysis of the formula C(n+r-1,r-1)?
Also I want to know how we can apply the above formula to find :
a) Number of solutions of the equation x+y+z= n (Where n is some natural number)
b) To find the number of terms in (a+b+c+d)^15
I want to understand the underlying concept.
Thanks
7 boxes numbered from 1-7 are arranged in a row.Each of them is filled with either black or blue colored balls such that no two adjacent boxes contain blue colored balls.In how many ways can the boxes be filled with balls
1.)23 2.)42 3.)34 4.)32 5.)33
@NishantGupta23 said: 7 boxes numbered from 1-7 are arranged in a row.Each of them is filled with either black or blue colored balls such that no two adjacent boxes contain blue colored balls.In how many ways can the boxes be filled with balls 1.)23 2.)42 3.)34 4.)32 5.)33
Is the answer = 34 ?
@[592846:YouMadFellow]: yes the answer is 34.Can u please tell me how u got it??
@NishantGupta23 said: 7 boxes numbered from 1-7 are arranged in a row.Each of them is filled with either black or blue colored balls such that no two adjacent boxes contain blue colored balls.In how many ways can the boxes be filled with balls 1.)23 2.)42 3.)34 4.)32 5.)33
Ans: 33 (Supposing that there will be atleast 1 box for Blue)
Case 1: If only 1 box contains Blue: 7 possibilities
Case 2: If 2 boxes contain Blue: 15 possibilities
Case 3: If 3 boxes contain Blue: 10 possibilities
Case 4: If 4 boxes contain Blue: 1 possibility
@[2199:muthuisfine] ans to c is 5^4 : 1st ball can be put into 5 boxes so 5 ways then 2nd in 5 ways similarly all 4 balls.
P.S.Correct me if i am wrong.
@NishantGupta23 said: @YouMadFellow: yes the answer is 34.Can u please tell me how u got it??
I labeled the boxes as B1, B2, B3... B7
For simplicity of symbols, I will take Black as Red(R) balls. Following are the cases for selection and arrangement:
(7R) = 1 way
(6R, 1B) = 7 ways
(5R, 2B) = 7!/(5!*2!) - 6!/5! = 15 ways
(4R, 3B) = 1 + 6 + 3 = 10 ways
(3R, 4B) = 1 way
total = 34 ways
For the case of (5R, 2B), consider all the arrangements (7!/(5!*2!)) but reduce the number of arrangements where the two blue balls are together
For the case of (4R, 3B), you can place the blue balls and put gaps in between, and place Red balls in the gaps in various ways ( 1 is when I consider 4 gaps, 6 is when I consider 3 gaps, 3 is when I consider 2 gaps)
Dont we have to consider the no of songs that can be put in the album? it can vary from 3 to 14.
@invAMARISH said: total no of song=14, Total no of way 2^12*5c1*6c1=2^12*30
@[557614:NishantGupta23]ans is 34. i) comb. 7 black =1, ii)comb. 6 black+1 blue=(6+1)c1 iii) comb. 5 black+2 blue=(5+1)c2 iv) comb. 4 black+ 3 blue=(4+1)c3 v) comb. 3black+4blue=(3+1)c4..no other comb. satisfies constraint..
sum= 1+7+15+10+1=34..
@[448836:sayantan1988] the arrangemnt is lyk _U_O_E_..so C,P,L can b arranged in 4 dashes in 4p3=4! ways..and for each such arrngmnt, U,O,E can b arranged among emselves in 3! ways..so correct ans= 4!*3!..:))
Please answer the following question:
A and B pick up a card at random from a well schufle pack of cards,one after the another,replacing it every time till one gets a heart.If A beins the game then the probability of B ends the game is :
1) 2/7 2)4/7 3)3/4 4)1/4 5)none of these
Please answer the following question:
A and B pick up a card at random from a well schufle pack of cards,one after the another,replacing it every time till one gets a heart.If A beins the game then the probability of B ends the game is :
1) 2/7 2)4/7 3)3/4 4)1/4 5)none of these
@NishantGupta23 said:Please answer the following question:A and B pick up a card at random from a well schufle pack of cards,one after the another,replacing it every time till one gets a heart.If A beins the game then the probability of B ends the game is :1) 2/7 2)4/7 3)3/4 4)1/4 5)none of these
Is it 5) none of these ?
I am getting 3/7 as the answer.
@[592846:YouMadFellow]:yes..can u tell how u got it??