# In how many ways can a batsman score a century in all $6$'s and $4$'s?

A batsman scored a century in all $6$'s and $4$'s. In how many ways can he do this?

The given answer is $8$, but there is no explanation, how are they doing it?

As century is $100$ runs a very loose translation of this problem would be "In how many ways we can get a sum of $100$ by using only $4$'s and $6$'s?"

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What is a century in this context? –  El'endia Starman Feb 18 '12 at 6:54
Abstract duplicate? –  The Chaz 2.0 Feb 18 '12 at 6:56
The question is about Cricket. It has nothing to do with Batman. –  Dan Brumleve Feb 18 '12 at 6:58
It's 1:00am on a weekend, @Dan... just being lighthearted :) –  The Chaz 2.0 Feb 18 '12 at 7:05
Count the number of positive integer solutions to $6x+4y = 100$ i.e. $3x+2y=50$ (assuming you want the batsman to hit at-least one six and assuming by century you mean an exact $100$ and not $102$ (or) $104$). Note that $x$ has to be even. Hence, we need $x$ to an even positive integer with $3x \leq 50$, which gives us $8$ options. –  user17762 Feb 18 '12 at 7:09

In ignorance of cricket, I will assume that a century means exactly $100$ runs.
Let's rephrase the question in terms of money. In how many ways can we have $100$ dollars in $4$ dollar bills and/or $6$ dollar bills? (It looks as if I don't know much about money either.)

The argument will be easier to grasp if we solve the equivalent problem of producing $50$ dollars in $2$ dollar and/or $3$ dollar bills. It is clear that we must use an even number of $3$ dollar bills, $0$ to $16$, and then we can make up the rest of the $50$ dollars with $2$ dollar bills. There are $9$ (not $8$) even numbers between $0$ and $16$ inclusive.

Note that if the order in which the types of scores were made matters, then the answer is hugely larger than $9$. Would you view $4$ then $6$ then $4$ as different from $4$ then $4$ then $6$?

Added: Derek Holt remarks that if one gets to $98$ with $4$'s and/or $6$'s, and then gets a $4$ or a $6$, one is still deemed to have scored a century with $4$'s and $6$'s. The same method as the one used above shows that there are $9$ ways to reach $98$. That interpretation gives an additional $18$ possibilities, for a total of $27$.

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Not points - they call them "runs". –  Gerry Myerson Feb 18 '12 at 7:56
@Gerry Myerson: Thanks, corrected. Is there a rule that brings the answer down to $8$? –  André Nicolas Feb 18 '12 at 8:08
There is no such rule, but I suppose you could interpret the question as implying that there has to be at least one 4 and at last one 6. That would bring the answer down to 8. Of course, in a real cricket match, if you reached 102 (or even 104 with the final mighty heave going for 6), you would still be deemed to have scored a century entirely with 4's and 6's. –  Derek Holt Feb 18 '12 at 10:15
In ignorance of cricket? You heathen! :-) –  Brian M. Scott Nov 5 '12 at 19:48

Using the generating function approach, the number of ways to score 100 runs, hitting only sixes and fours, where the order does not matter, is the coefficient of $x^{100}$ in the expansion of:

$$(1+x^4+x^8+x^{12}+x^{16}+\dots+x^{92}+x^{96}+x^{100})(1+x^{6}+x^{12}+x^{18} + \dots+ x^{90}+x^{96}).$$

Walpha shows that the answer is, indeed, nine.

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One way to score a century is to score $4$ runs twenty-five times. You can replace three $4$s with two 6s a number of times, that number being from $0$ to $\lfloor 100/12 \rfloor = 8$, with an extreme case of two $4$s and sixteen $6$s.

So that gives nine ways, as André Nicolas says.

Perhaps scoring "a century in all $6$s and $4$s" carries the implication that at least on $6$ and at least one $4$ are scored. So the number is instead from $1$ to $8$, and there are eight ways.

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If you want exactly $100$ (and not $102$ or $104$, etc) Then these are the possibilities $10\cdot 6+10\cdot 4=100$, $12\cdot 6+7\cdot 4=100$, $14\cdot 6+4\cdot 4=100$, $16\cdot 6+1\cdot 4=100$, $8\cdot 6+13\cdot 4=100$, $6\cdot 6+16\cdot 4=100$, $4\cdot 6+19\cdot 4=100$, $2\cdot 6+22\cdot 4=100$ And if you allow that no $6$'s required and he can make $100$ with only $4$'s then this is the $9$th possibility, $0\cdot 6+25\cdot 4=100$.

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For some basic information about writing math at this site see e.g. here, here, here and here. –  Julian Kuelshammer Nov 5 '12 at 19:34