2022 AMC 12A Problems/Problem 24

Problem

How many strings of length 5 formed from the digits 0, 1, 2, 3, 4 are there such that for each , at least of the digits are less than ? (For example, 02214 satisfies this condition because it contains at least 1 digit less than 1, at least 2 digits less than 2, at least 3 digits less than 3, and at least 4 digits less than 4. The string 23404 does not satisfy the condition because it does not contain at least 2 digits less than 2.)

Solution

Denote by the number of -digit strings formed by using numbers , where for each , at least of the digits are less than .

We have the following recursive equation: and the boundary condition for any .

By solving this recursive equation, for and , we get \begin{align*} N \left( p , 1 \right) & = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\ & = \sum_{i = 0}^{p - 1} \binom{p}{i} \\ & = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\ & = 2^p - 1 . \end{align*}

For and , we get \begin{align*} N \left( p , 2 \right) & = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\ & = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) - \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right) - p \\ & = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\ & = 3^p - 2^p - p . \end{align*}

For and , we get \begin{align*} N \left( p , 3 \right) & = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\ & = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) - \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} \left( p - i \right) \right) - \frac{3}{2} p \left( p - 1 \right) \\ & = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p - \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1} - \frac{3}{2} p \left( p - 1 \right) \\ & = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) . \end{align*}

For and , we get \begin{align*} N \left( 5 , 4 \right) & = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\ & = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\ & = \boxed{\textbf{(E) 1296}} . \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/mj78e_LnkX0

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)