@YouMadFellow not in options bro
@The_Loser said:in a row of 20 seats , in hw many ways can 3 blocks of consecutive seats wid 5 seats in each block be selected??Ans
A,B,C - Blocks of 5 consecutive seats
g1 A g2 B g3 C g4
g1+ g2+ g3 + g4 = 5 (g1,g2 etc are gaps between those selections)
=> 8C3 = 56 ways ?
six people are qued up at a theatre box office three of them have only 5 rupee and remaining 3 10 rupee notes.there is no cash in the box office when it opens and price of each ticket is 5 rupee.find the chance that no customer will have to wait?
@jaiswalicsi said:In how many ways 4620 pizzas can be divided equallty among n people such that they cant furthur divide equally among n people.
@The_Loser said:Each side of a cube is painted red or blue or green. How many distinct colour patterns possible?
1/24(n^6+3n^4+12n^3+8n^2)
=1/24(729+243+324+72)
=57
=1/24(729+243+324+72)
=57
@jaiswalicsi said:six people are qued up at a theatre box office three of them have only 5 rupee and remaining 3 10 rupee notes.there is no cash in the box office when it opens and price of each ticket is 5 rupee.find the chance that no customer will have to wait?
5 10 5 10 5 10 - 36
5 10 5 5 10 10 - 36
5 5 5 10 10 10 - 36
5 5 10 5 10 10 - 36
5 5 10 10 5 10 - 36
180/720=1/4
@The_Loser said:in a row of 20 seats , in hw many ways can 3 blocks of consecutive seats wid 5 seats in each block be selected??Ans
_xxxxx_xxxxx_xxxxx_
xxxxx, xxxxx, xxxxx are the 3 blocks of 5 consecutive seats
a1 in first gap, a2 in 2nd, 3 in 3rd and a4 in last
a1 + a2 + a3 + a4 = 5
C(8, 3) = 56 ways
@jaiswalicsi said:six people are qued up at a theatre box office three of them have only 5 rupee and remaining 3 10 rupee notes.there is no cash in the box office when it opens and price of each ticket is 5 rupee.find the chance that no customer will have to wait?
Application of Catalan Number
C(6, 3) - C(6, 2) = 5 ways
Probability = 5/C(6, 3) = 1/4
@krum said:4620=2^2*3*5*7*112*2*2*2*2=3231edited
What do we need to find here ? Possible values of n, right ?
n can be any of the following:
1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 14, 15, 20, 21, 22, 28, 30, 33, 35, 42, 44, 55, 60, 66, 70, 77, 84, 105, 110, 132, 140, 154, 165, 210, 220, 231, 308, 330, 385, 420, 462, 660, 770, 924, 1155, 1540, 2310,4620
But for n = 1,2 (the resultant will again be divisible by 1 and 2) , Rest all cases we cannot divide them again
Ex: 12 divides 4620 => Each person will get 385 pizzas, now these 385 pizzas cannot be divided into 12 people again !
=> If n divides 4620, n^2 should not divide 4620
What is the mistake here ?
@krum said:5 10 5 10 5 105 10 5 5 10 105 5 5 10 10 105 5 10 5 10 10 5 5 10 10 5 105 ways?
dont have oa but i did like this
total number of arrangements =6!/3!*3! =20 (as it is arragmnt of money amount)
if a person comes with 10 rs first then whole case is distrtd
so un favorable ways=10__ __ __ __ __(5!/3!*2!)=10 cases
favourable=10 cases
prob =1/2
please check this
total number of arrangements =6!/3!*3! =20 (as it is arragmnt of money amount)
if a person comes with 10 rs first then whole case is distrtd
so un favorable ways=10__ __ __ __ __(5!/3!*2!)=10 cases
favourable=10 cases
prob =1/2
please check this
@jaiswalicsi said:dont have oa but i did like thistotal number of arrangements =6!/3!*3! =20 (as it is arragmnt of money amount)if a person comes with 10 rs first then whole case is distrtd so un favorable ways=10__ __ __ __ __(5!/3!*2!)=10 cases favourable=10 cases prob =1/2please check this
you missed some cases like
5, 10, 10, _ _ _ 3 cases
5, 10, 10, _ _ _ 3 cases
5, 10, 5, 10, 10, 5 - 1 case
5, 5, 10, 10, 10, 5 - 1 case
So, 10 + 5 case should be removed
Boxes numbered 1,2,3,4 and 5 are kept in a row and they which are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?
(a) 8
(b) 10
(c) 15
(d) 22
(a) 8
(b) 10
(c) 15
(d) 22
@gautam22 said:sir 47 hona chahiye.....1 consider nahi hoga ye 31 kyun kara hai aapne.....48-2(factor hai jab)....47 hua.......4620-nk/n.....sirf 2 ke case mein dobara divisible ho hayega to sirf vahi hi hatayenge............plz correct me if i m wrong
1 is also there. 46, then ?
@joyjitpal said:Boxes numbered 1,2,3,4 and 5 are kept in a row and they which are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then, how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?(a) 8 (b) 10 (c) 15 (d) 22
I think answer should be 22
For 0 blue balls---> 1 ways
For 1 blue ball-->5 ways
For 2 blue balls--->15 ways (5 ways for one and 3 ways for the other)
For 3 blue balls---> 1 way
For 4--->not possible
@gautam22 said:1 se divide 4620 aa gaye ek ke paas....fir divide kara to zero aaya remainder poora divide ho to gaya............kya galat kar raha hoon sir??
No, I mean if n = 1 => When we divide 4620 among 1 people, he will get 4620, then if he divides it again among 1 people => (the other person will again get 4620) => It can be divided equally ...
=> n = 1 and 2 should not be considered => 48 - 2 = 46
@gautam22 said:other person will again get ka matlab samjh nahi aaya sir 4620 gaya bacha zero.....other person ka matlab sir?
Suppose chill sir gives me 4620 pizzas 
...Now, since n = 1 here -> can I again divide it equally among 1 person , Answer is Yes .. So, I can give it to YOU !
...Now, since n = 1 here -> can I again divide it equally among 1 person , Answer is Yes .. So, I can give it to YOU ! Why are you looking at the remainder ? We need to look at the quotient ! 😐 , because that's what every person gets
Suppose 2 people are there => 4620/2 = 2310 pizzas for each of them, now each of them can divide 2310 equally among 2 people (because 2310/2 = 1155 pizzas) => So, this is a rejected case
@joyjitpal said:@chillfactor what is catalan number?plz explain
check this link, I don't think i can explain better