2
$\begingroup$

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?"

  • 4
    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.2012-02-18

4 Answers 4

4

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 $27$.

  • 0
    In ignorance of cricket? You heathen! :-)2012-11-05
0

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.

0

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.

0

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, $0\cdot 6+25\cdot 4=100$.

  • 1
    For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex).2012-11-05