Is this solvable ??

Is this solvable ??

Find no of ways by which we can stack 12 coins in a line so that each coin lies on table or on 2 coins

12 Answers

1
Philip Calvert ·

?
how can a coin only lie on two coins cuz then the coin below will be lying on one coin

24
eureka123 ·

like this...here black coins are on table..and grey one is placed on top

1
Philip Calvert ·

ok

1
taran ·

6C5 + 7C4 +8C3 + 9C2 +10 C1 +11C0

24
eureka123 ·

final ans ?

1
taran ·

154 summing as written above

24
eureka123 ·

no..


WARNING FOR ALL..DONT SEE THIS WITHOUT TRYING

ans is 234

1
Avinava Datta ·

are all the coins identical ??

1
taran ·

i had assumed so

1
Amritansh Bharech ·

the ans. if u do
6C5 + 7C4 +8C3 + 9C2 +10 C1 +11C0
is 144 and not 154
and that seems to be the right ans. if the coins r identical

1
Samarth Kashyap ·

i'm not getting the rite ans...i'm getting it as 274......i guess i've counted many combinations twice.....somebody pls correct me.....but , here's my approach
we v 12 coins....we place some of them in a line and the remainig in our hands, we place them on the coins on the table so as to satisfy the given condition....

so the combinations for (line,hand)={(12,0),(11,1)........(5,7)}...it ends here....u cant hv 4 on the line & then place the remaining

(12,0)-1
(11,1)-10
(10,2)-9C2
(9,3)-8C3+7
(8,4)-7C4+5*2+6*1*5
(7,5)-6C5+3*3+4+5*4C2
(6,6)-4
(5,7)-1+2

total=274
here's an illus for (7,5)

1
Amritansh Bharech ·

this approach is wrong i think bcoz u cannot place a coin in a third row
then that coin does not lie on 2 coins but more than 2 coins

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