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?"
 A: 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 $27$. 
A: 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.
A: 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.
A: 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, 
$0\cdot 6+25\cdot 4=100$.
