Question

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A soccer team has 22 available players. A fixed set of 11 players starts the game, while the other 11 are available as substitutes. During the game, the coach may make as many as 3 substitutions, where any one of the 11 players in the game is replaced by one of the substitutes. No player removed from the game may reenter the game, although a substitute entering the game may be replaced later. No two substitutions can happen at the same time. The players involved and the order of the substitutions matter. Let n be the number of ways the coach can make substitutions during the game (including the possibility of making no substitutions). Find the remainder when n is divided by 1000.

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Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

122 is the answer of n mod 1000.

$\huge\star{\green{\underline{\mathfrak{Explanation :-}}}}$

Given :-

Available players = 22

Number of players starting the game = 11

Substitutes = 11

Coach can only make 3 substitutions or even no substitution which means 0, 1, 2 and 3 are the number of substtuitions that can be made.

To find :-

If n is the number of ways the coach can make substitutions during the game then....

We need to find out the remainder of n is divided by 1000 that is n mod 1000.

Solution :-

Let's solve it by casework.

Case 1 :

Let the number of substitutions made be 0.

So, the number of ways 0 substitution can be done is 1 because all the same 11 players will play the soccer.

So, for Case 1 let the final answer be 1.

Case 2 :

Let the number of substitutions made be 1.

So, the coach has 11 choices to remove the players from the players that are playing and another 11 choices to substitute instead of the removed one's.

So, $11\times 11$

$= 121 \: ways$

So, final answer in Case 2 = 121.

Case 3 :

Let the number of substitution made is 2.

Now, here the substitution has to be done 2 times.

In 1st time : The number of choices of players removed can be 11 and the choices of the number of players substituted instead of the removed one's can be 11.

And in the 2nd time : The number of choices of the players removed is still 11 and the number of choices of the players substituted instead of the removed players will be 10 because the one removed in the 1st time can not join the playing team again.

So, the number of ways to make 2 substitutions are : $11 \times 11\times 11\times 10$

So, the final answer here is :- 13310.

Case 4 :

Now, let the number of substitution made is 3.

So, the substitution will occur 3 times.

In the 1st time : The number of choices of the players removed is 11 and the number of choices of the players substituted instead of the removed one's is 11.

In the 2nd time : The number of choices of the players removed is still 11 and the number of choices substituted instead of thr one's removed becomes 10 because the one removed in the first time can not re enter the game.

In the 3rd time : The number of choices of the players removed will be 11 again and the number of choices of the players substituted will become 9 because the one's removed in first and second time can not join the playing team again.

So, the number of ways 3 substitutions can made are : $11 \times 11 \times 11 \times 10 \times 11 \times 9$

So, the final answer is = 1,317,690.

_____________

Now, the total number of ways the coach can make substitutions during the game (including the possibility of making no substitutions) = n :-

(We will get n by adding all the final answers of all the cases above)

$\leadsto{1 + 121 + 13310 + 1,317,690 }$

$\leadsto {1 331,122}$

Now, n = 1,331,122

We need to find n mod 1000.

That is 1,331,122 mod 1000

When we will divide 1,331,122 by 1000 ; we get the quotient as 1331 and the remainder as 122.

So, 1,331,122 mod 1000 is 122.

Final Answer :- 122.

_________________

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_________________

Answer 2

Answer:

The answer is $\boxed{\bf 122}}$

Step-by-step explanation:

Given Information :

Total Available players - 22

Substitutes = 11

Number of players starting the game = 11

To Find :

The remainder when n is divided by 1000.

Solution :

This question can be solved by two methods :

• Recursion.
• Casework.

First I should explain the method "Recursion"

In this method,

The number of ways to sub any number of times must be multiplied by the previous number.

So,

There are 0 - 3 substitutions. The situation for 0 subs is one.

Now,

The ways to identify after n subs is the product of the number of new subs (12-n).

The players that can or can't be ejected is 11.

We know that :

The formula of n.

So,

$\sf\\ \\\implies a_n=11(12-n)a_{n-1} \\\\\implies a_0=1$

Adding from 0 to 3 :

$\sf \\ \implies 1+11^2+11^{3}\cdot 10+11^{4}\cdot 10\cdot 9\\ \\\implies 10+9\cdot11\cdot10=10+990=1000$

$\implies 1+11^2+11^3\cdot (10+11\cdot9)= 1+11^2+11^3\cdot (1000) .$

_______________[ Re-aranged! ]

While considering modulo as 1000.

This is left,

$\implies \sf 1+11^2=\boxed{122}$

_________________________

Note that, this question can be solved by the method - "CaseWork"

$\rule{300}{1.5}$

#AnswerWithQuality.

#BAL! :)