ISI 2004 MIII

Question

In how many ways can three person, each throwing a single die once, make a score of $11$

• A. $22$
• B. $27$
• C. $24$
• D. $38$

Comments

This is combination with limited repetation

Answer 1

We can solve it by star and bar method.

Suppose $1$st person's score is $x_1.$

Suppose $2$nd person's score is $x_2.$

Suppose $3$rd person's score is $x_3.$

We want $x_1+x_2+x_3 = 11$ with constraints $1\leq x_1\leq 6$ , $1\leq x_2\leq 6$ , $1\leq x_2\leq 6$

So $x_1,x_2,x_3$ have minimum values $1$ and remaining $8$ can be scored as $x_1+x_2+x_3 = 8.$

Total no. of ways to score $8=\binom{3+8-1}{2} = \binom{10}{2}.$ 

Now we have to subtract the ways in which either $x_1$ or $x_2$ or $x_3$ have values $\geq 7 .$

When $x_1 \geq 7,$ we get $x_1+x_2+x_3=2$ (we subtract $6$ from RHS because $x_1$ is already assumed to be $\geq 1$ and  $6+1=7$).

No. of ways $= \binom{3+2-1}{2} = \binom{4}{2}$

Doing same for $x_2$ and $x_3.$

Note: Only one of the $3$ can have score $\geq 7$ as if 2 suppose $x_1$ and $x_2$ both is $7$ then total will exceed $8$.

So, total no. of ways = $\binom{10}{2} - \left [ \binom{4}{2} +\binom{4}{2} + \binom{4}{2} \right ] = 27.$

(B) is the correct answer.

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Another way - 
Contraints - $1\leq x_1,x_2,x_3\leq 6.$
Now, 11 can be broken as - 
$\{6,4,1\} \rightarrow$ can be scored in $3! \ ways$
$\{6,3,2\} \rightarrow$ can be scored in $3! \ ways$
$\{5,5,1\} \rightarrow$ can be scored in $\frac{3!}{2!} \ ways$
$\{5,4,2\} \rightarrow$ can be scored in $3! \ ways$
$\{5,3,3\} \rightarrow$ can be scored in $\frac{3!}{2!} \ ways$
$\{4,4,3\} \rightarrow$ can be scored in $\frac{3!}{2!} \ ways$   

$3! + 3! +\frac{3!}{2!} + 3! + \frac{3!}{2!} + \frac{3!}{2!}$

$\Rightarrow 3.3! + 3. \frac{3!}{2!} = 18+ 9 =27$

Comments

@Na462 @shaik sir There is nothing special in it. Anyway thanks, both of you :)

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@Soumya29

why r u taking $x_{1}\geq 7$?

because we already have taken $1\leq x_{1}\leq 6$

right?

then why again taking different value?

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Because first we solved for >= 1 I.e. the atleast case now to bound it till 6 I.e. <= 6 we need to subtract all the cases that involves >= 7 values. Now the result is between 1 and 6 both included. :)

Answer 2

The sum 11 can be broken as 6 + 5

Let us name the players as A, B and C.

Case 1: Fix 6 and break 5

Suppose A throws 6. Players B and C can throw dice in following four ways to make sum 5

(1,4), (2,3), (3,2) and (4,1)

As any of the three players can throw value 6, so there are total $3\times 4=12$

Case 2: Fix 5 and break 6

Suppose A throws 5. Players B and C can throw dice in following five ways to make sum 6

(1,5), (2,4), (3,3), (4,2), (5,1)

As any of the three players can throw value 5, so there are total $3\times 5=15$

Adding case 1 and case 2 there are total $12 + 15 = 27$ ways

Comments

Your case 3 has repetitions of the pair (5,1,5) because in this case both A and C have already thrown 5 and interchanging those 5s will not add another combination to the answer. Therefore fixing 5, you get only 3x4= 12 ways and you missed out the combination of (4,4,3) there are 3 ways to throw 4,4 and 3. The answer still remains as

12(case 1) + 12(case 2) + 3(4,4,3) = 27

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Yes.. missed out the repetitions in case 2 where the triples (5,1,5), (5,5,1) , (1,5,5) get counted twice. So case 2 gives 15-3 =12 ways

And case 3 of combination (4,4,3) gives 3 more ways.

So total remains 27.

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Yes correct

Answer 3

x1 + x2 + x3 = 11  1<=xi<=6

find coeff of [ x^11 ] in  [( x + x^2 + x^3 + x^4 + x^5 + x^6)]^3

=>   x^3  [(  1 + x + x^2 + x^3 +  x^4 + x^5)]^3

=>  find coeff of [ x^(11-3) ] in  [(  1 + x + x^2 + x^3 + x^4 + x^5 )]^3

=>  find coeff of [ x^8 ] in  [(  1 - x^6 ) /  (1 - x_)]^3

=>  find coeff of [ x^8 ] in  ( 1 - x^6 )^3 *  (1 - x)^ -3

=> [$_{0}^{3}\textrm{C}$ *   $_{8}^{-3}\textrm{C}$ ]  -  [$_{1}^{3}\textrm{C}$ *   $_{2}^{-3}\textrm{C}$ ]

=> [$_{0}^{3}\textrm{C}$ *   $_{8}^{10}\textrm{C}$ ]  -  [$_{1}^{3}\textrm{C}$ *   $_{2}^{4}\textrm{C}$ ]

=>  45 - 18 = 27 ways

Comments

How do u calculated it ??

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for basics watch all videos of trevtutor

 https://www.youtube.com/watch?v=DZFS3GHSc88&list=PLDDGPdw7e6Aj0amDsYInT_8p6xTSTGEi2&index=13

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Nice explanation dude

Answer 4

An answer using generating functions:

The output of every roll can be either of 1,2,3,4,5,6.This sequence can be modelled by the expansion $x+x^2+x^3+x^4+x^5+x^6$.

This is a Geometric progression with common ratio = x.

$ x(1+x+x^2+x^3+x^4+x^5)$

The generating function for this sequence is $x.\frac{1-x^6}{1-x}$. Since  there are three dices to be rolled, hence we have the required expression as :$\left ( x.\frac{1-x^6}{1-x} \right )^{3}$.

It can be re-written as $ x^3\left (\frac{1-x^6}{1-x} \right )^{3}$.

Multiplication by $x^3$ means that every power of x is shifted by 3 positions.So, effectively we need to find out the coefficient of $x^8$ from  $ \left (\frac{1-x^6}{1-x} \right )^{3}$.

Now we need the coefficient of $x^{8}$ out of this generating function.

 $=\left ( \frac{1-x^6}{1-x} \right )^{3}=(1-x^6)^3(1-x)^{-3}$

$=(1-x^{18}-3x^6(1-x^6))(1-x)^{-3}$

$=(1-x^{18}-3x^6+3x^{12})(1-x)^{-3}$

x¹⁸ and x¹² are of no use to us because we need terms that will add up to 8 and  these terms are already greater than 8.

If we take 1 from first bracket then we need whole $x^8$ from 2nd bracket and the coefficient of $x^8$ in 2nd expression is = $\binom{3+8-1}{8}$.

If we take -3x⁶ from 1st bracket ,then we need coefficient of $x^2$ from 2nd bracket.The coefficient of $x^2$ in 2nd bracket is $\binom{3+2-1}{2}$.Hence the coefficient=$(-3)(6)=-18$

Therefore the final coefficient of $x^8$ is $\binom{3+8-1}{8} - 18=27$

NOTE: Those who are not able to understand coefficient extraction from 2nd bracket should refer the extended binomial theorem.