When 7 fair standard 6-sided dice are thrown, the probability that the

Question

When 7 fair standard 6-sided dice are thrown, the probability that the sum of the numbers on the top faces is 10 can be written as  \frac{n}{6^7} , where n is a positive integer. What is n?

(A) 42
(B) 49
(C) 56
(D) 63
(E) 84

Archit3110 · Answer

total possiblities ;

4,1,1,1,1,1,1: 7 ; 7!/6!

3,2,1,1,1,1,1: 42; 7!/5!

2,2,2,1,1,1,1: 35; 7!/3!*2! 
sum ; 7+42+35; 84
IMO E

exc4libur · Answer

7 dices that face sum 10:
  1/6^7 * 7!/6!1! (ways to arrange  )
  1/6^7 * 7!/5!1!1! (ways to arrange  )
  1/6^7 * 7!/4!3! (ways to arrange  )
probability: (7+42+35)/6^7 = 84/6^7

ans (E)

nick1816 · Answer

Question is similar to how many ways we can write 10 as the sum of 7 positive integers.

a|b|c|d|e|f|g

Number of ways = (6+3)C3 = 9C3 = 84

AnirudhaS · Answer

Sorry  nick1816  would you dumb the highlighted portion down for me please?

nick1816 · Answer

a, b, c, d, e,f, g will take minimum value equals to 1; hence, you have to split remaining (10-7) = 3 at the 7 places

A | B | C | D | E | F | G

You have to arrange 6 similar dashes ( |) and 3 identical units.

Total number of ways =  (6+3)!/6!3!  = 9C3.

Or you can just remember the formulae for the  positive integral solutions  for the following equation

a_1+a_2+a_3 +.....+ a_r = n

total possible solutions = (n-1) C (r-1)

In this question n= 10 and r = 7

sujal998 · Answer

total 7 dice so 
1 1 1 1 1   1 4   1 combination will get 7c2 ways 7*6/2*1 =21 
1 1 1 1 1   2 3   
1 1 1 1 1  4 1 
1 1 1 1 1   3 2

4*21=84
ans E

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When 7 fair standard 6-sided dice are thrown, the probability that the

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