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?"
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 $27$.
Consider the number of $4$s $= x$, number of $6$s $= y$. Then the total score should be $100$.
Therefore, $4x+6y=100$. Solving this we get, $x=25-3/2(y)$
Here $y$ should be integer and it should be even. The range of $y$ will start from $0$ to $16$. It means $y$ will be $0,2,4,6,8,10,12,14,16$. Total $9$ Combinations will come.
Note: $y$ should not be more than $18$, because if we take $x$ value become negative. It won't be possible.
Hence, the answer is $9$.
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.
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.
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, $0\cdot 6+25\cdot 4=100$.
Using only 6s, the batsman can reach 96 in 16 balls and then hits a 4 on the 17th ball to get to 100.
Using only 4s, he can do that in 25 balls.
Considering there is no dot balls, he can achieve his century in 17 to 25 balls
I was thinking to be lazy and just subtract the numbers to get 9 but lets try to think it through
2 balls of 6s can be replaced by 3 balls of 4s and give the same 12 runs. So 17 - 2 + 3 would give us a possible combination of 4s and 6s (namely 14 6s and 4 4s) and similary another -2+3 would give 19 ball combination with 12 6s and 7 4s and you can go ahead in successions of 1 ball to get a proper combination for this context. Hence there are possible and allowed combinations in 17 to 25 balls. Therefore there are 25 - 17 + 1 = 9 ways.
