P&C- a doubt from the Dark Knight

The no. of ways in which a score of 11 can be made from a throw by 3 persons, each throwing a die once is :

A)45
B)18
C)27
D)68

Ans : C

NO USE OF MULTINOMIALS PLEEEEEEAAAASE

11 Answers

1
Bicchuram Aveek ·

Anyone watching it ?????

62
Lokesh Verma ·

if you think closely, this is the same as teh number of ways to get 7 by the throw of dice by 3 ppl

This will be a bit dirty counting.. but i hope this observation makes the question slightly simpler .. (at least psychologically :D)

1
Bicchuram Aveek ·

Couldn't get it, Sir . :-(

49
Subhomoy Bakshi ·

bhaiya.....i also cudn't get it....

62
Lokesh Verma ·

so the number of ways to get 11

for each dice get 7-i instead of i

so there is a one one onto relation between the number of ways you can get 11 and the number of ways you can get 21-11 = 10

(sorry hadbadi me gadbadi :P)

1
Unicorn--- Extinct!! ·

"hadbadi me gadbadi" apka fav dialogue hai?[6]

62
Lokesh Verma ·

nahi.. but the most common thing i do here [6]

49
Subhomoy Bakshi ·

still not got....

1
Bicchuram Aveek ·

Sir mera Sir ke upar se Tangent ho kar nikal gaya

1
Bicchuram Aveek ·

Nishant Bhaiyaaaaaaaaaaaaaaaaaaa

calling for help.......!!!!!!

106
Asish Mahapatra ·

The no. of ways in which a score of 11 can be made from a throw by 3 persons, each throwing a die once is

x11 in (x1+x2 + ..+x6)3

= x8 in (1-x6)3(1-x)-3

= x8 in (1-3x6)(1-x)-3

= 10C8 -3*4C2

= 45 - 18

= 27

Now for non-multinomial method

suppose u fix that one person gets 6. then the other two persons can get (1,4) (2,3) so total no. of ways = 2*3!

Now let one person get 5. Then other two can get (1,5) (2,4) (3,3) so total no. of ways = 2.3!/2! + 1.3!

Now let one person get 4. Then other two can get (1,6) (ignored as already counted) (2,5) (ignored as already counted) (3,4) so total no. of ways = 3!/2!

So total = 27

Now why havent i considered when one person gets 3 or 2 or 1?

IT is because if one gets 3 then others can get (2,6) (3,5) (4,4) which are all considered above.

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