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"A hockey team consists of 1 goalkeeper, 4 defenders, 4 midfielders and 2 forwards. There are four substitutes: 1 goalkeeper, 1 defender, 1 midfielder and 1 forward. A substitute may only replace a player in the same category e.g. midfielder for midfielder. Given that a maximum of 3 substitutes may be used and that there are still 11 players on the pitch at the end, how many different teams could finish the game?"

The solutions say that if 4 substitutes are allowed, 2×5×5×3=150 different teams could finish the game. But 1×4×4×2=32 of those substitutions require four substitutions, so the answer is 118.

Could you please explain the logic with which '32' is calculated?

Shaun
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Casper C.
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    Firstly, the number $2\times 5\times 5\times 3$ counts the number of ways that you could have one of the goalkeepers, one of the defenders, etc... off the field on the bench at the end of the game. Alternatively worded, it would be $\binom{2}{1}\binom{5}{4}\binom{5}{4}\binom{3}{2}$ to count the number of ways that $1,4,4,2$ of each position are on the pitch at the end respectively. Once a player is substituted out it cannot return to the field. Using all four substitutes then would require each of the substitutes to be used. For each category, pick who it was who was taken off the pitch. – JMoravitz Aug 04 '17 at 19:02
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    Here's a MathJax tutorial :) – Shaun Aug 04 '17 at 19:03
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    Pick which one of the $1$ original goalkeepers was taken off and replaced by a substitute goalkeeper. Pick which one of the $4$ original defenders was taken off and replaced by a substitute defender. Etc... Apply multiplication principle to count the number of such arrangements by multiplying the number of options available at each step giving $1\times 4\times 4\times 2=32$ such arrangements where each position got a substitute. As these were the "bad" arrangements from those we counted before, we subtract this from the original total giving $150-32=118$ valid arrangements with$\leq 3$ subs – JMoravitz Aug 04 '17 at 19:03
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    As an aside, I highly discourage referring to such arrangements as "permutations" when they are not in fact explicitly permutations. Call them something more general such as "arrangements" or "outcomes." A permutation would specifically require every available object be included in the final arrangement exactly once and order within the arrangement matters. – JMoravitz Aug 04 '17 at 19:07

2 Answers2

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No. of teams allowing 4 substitutions at a time = $1C1 * 4C1 * 4C1 * 2C1 = 1*4*4*2=32$

NOTE: No. of ways r items can be selected from n available distinct objects = $nCr$

Nirvana Guha
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There is a total of 2 GK, 5 DEF, 5 MID, 3 ST. (including all reserves).

Are you with me so far?

Good.

Now, out of the total, how many ways are there to select a GK?

The answer is simple - 2 possibilities. This can be expressed as $\binom{2}{1}$ or the number of ways to select 1 person from 2.

Now, similarly, how many ways are there to select a DEF?

The answer is simple - 5 possibilities. This can be expressed as $\binom{5}{4}$ or the number of ways to select 4 persons from 5.

And so on...

Now, all these combinations are set together to form the team.

You follow?

So the total number of teams that can be formed is given by $\binom{2}{1}*\binom{5}{4} *\binom{5}{4}*\binom{3}{2} = 150$

Now, out of these, many cases contain 4 substitutions, which is not allowed as per the rules of THIS game. Only 3 are allowed.

Our next task is to eliminate all such cases.

How many such cases are there in total?

1 for GK

4 for DEF

4 for MID

2 for ST

(following previous logic)

Since all these occur at the same time, we have $1*4*4*2=32$ extra cases which are illegal.

So, the required answer is $150-32=118$

Please let me know if it helped.

Nirvana Guha
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  • also note that once a player is taken off the field, he cannot be re-substituted. – Nirvana Guha Aug 04 '17 at 20:27
  • Thank you very much for this. How, logically, do we conclude that there are 4 cases where the substitution of the midfielder required four substitutions and two cases for strikers etc.? That's the part I don't understand. Thank you for your help. – Casper C. Aug 04 '17 at 23:03
  • ok, let me see if i can clarify this.

    i think one example will be sufficient. take the MID players for instance. assume that the 4 opening players in the game are Bob, Joe, Billy, and Sandy. We have Rob on the bench, waiting as the reserve/substitute.

    In how many ways can Rob actually replace a player on the opening team? He can replace one of Bob, Joe, Billy, and Sandy. He can't replace anyone else, they aren't midfielders. So out of these four, (Bob, Joe, Billy, and Sandy) he can select only 1 player. Hence, $\binom{4}{1}$ or simply put, 4 ways to substitute.

    Did this help?

     – Nirvana Guha Aug 06 '17 at 13:47