# How many different license plates start with the letter A if letters and digits cannot be repeated? Enter the exact numeric answer.

a) Consider the equation $x_1 + x_2 + x_3 + x_4 = 35$. How many different solutions does this equation have if all the variables must be positive integers? Enter the exact numeric answer.

b) Suppose that a license plate consists of three letters followed by three digits. How many different license plates start with the letter A if letters and digits cannot be repeated? Enter the exact numeric answer.

My work

a) $C(38,35)$

b)$1*P(26,2)*P(10,3)$

A_ _ _ _ _ . The last three number should not be repeated, so $P(10,3)$ and then the letters can be chosen randomly so $C(26,2)$

• To chose the letters, you don't want $C(26,2)$, because order matters. Since one letter is already used, you want $P(25,2)$. – G Tony Jacobs Apr 24 '18 at 23:57

(a) Almost. The restriction is that all variables must be positive integers, so we can rewrite this as

$$y_1+y_2+y_3+y_4 = 31$$

with $y_i = x_i-1$ being a non-negative integer for all $i$. Then stars-and-bars gives us $C(34, 31) = C(34, 3)$.

(b) Almost. As Tony Jacobs points out in the comments, the two letters should be represented by $P(25, 2)$, since you have to avoid $A$ (leaving only $25$ letters left to chose from), and order matters, as it does with the digits.

• @tienlee: Careful: Were the numbers in that case required to be positive, or merely non-negative? (In other words, could $x_1 = 0$?) – Brian Tung Apr 25 '18 at 0:06
• But, my question was $x1+x2+x3+x4=35$ and I am getting the answer as $(38,35)$ – tien lee Apr 25 '18 at 0:07
• @tienlee: Suppose you needed to answer $x_1+x_2+x_3 = 5$ with positive integers. Should the answer be $C(7, 5)$ or $C(4, 2)$? Enumerate the solutions—how many are there? – Brian Tung Apr 25 '18 at 0:26
• The solution to $y_1+y_2+y_3+y_4 = 31$ (with non-negative $y_i$) is a solution to $x_1+x_2+x_3+x_4 = 35$ (with positive $x_i$) when you take $x_i = y_i+1$. – Brian Tung Apr 25 '18 at 0:27

Hint for a)

Write $$35=1+1+\dots+1$$ Now, in how many ways can you place three commas? For instance $1,1+1,1+1+1,1+\dots+1$.

For b), you are not considering that the digits come after the letters.

• The OP has taken into account the positions of the letters and numbers. The OP's mistake is that the letter A has already been used, so there are only $25$ letters from which to choose the remaining letters. – N. F. Taussig Apr 25 '18 at 0:10