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| − | ==Problem==
| + | #redirect [[2022 AMC 10A Problems/Problem 24]] |
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| − | How many strings of length <math>5</math> formed from the digits <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math> are there such that for each <math>j \in \{1,2,3,4\}</math>, at least <math>j</math> of the digits are less than <math>j</math>? (For example, <math>02214</math> satisfies this condition
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| − | because it contains at least <math>1</math> digit less than <math>1</math>, at least <math>2</math> digits less than <math>2</math>, at least <math>3</math> digits less
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| − | than <math>3</math>, and at least <math>4</math> digits less than <math>4</math>. The string <math>23404</math> does not satisfy the condition because it
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| − | does not contain at least <math>2</math> digits less than <math>2</math>.)
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| − | <math>\textbf{(A)} \, 500 \qquad\textbf{(B)} \, 625 \qquad\textbf{(C)} \, 1089 \qquad\textbf{(D)} \, 1199 \qquad\textbf{(E)} \, 1296 </math>
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| − | ==Solution (Recursive equations approach)==
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| − | Denote by <math>N \left( p, q \right)</math> the number of <math>p</math>-digit strings formed by using numbers <math>0, 1, \cdots, q</math>, where for each <math>j \in \{1,2, \cdots , q\}</math>, at least <math>j</math> of the digits are less than <math>j</math>.
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| − | We have the following recursive equation:
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| − | <cmath>
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| − | N \left( p, q \right)
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| − | = \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1
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| − | </cmath>
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| − | and the boundary condition <math>N \left( p, 0 \right) = 1</math> for any <math>p \geq 0</math>.
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| − | By solving this recursive equation, for <math>q = 1</math> and <math>p \geq q</math>, we get
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| − | <cmath>
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| − | \begin{align*}
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| − | N \left( p , 1 \right)
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| − | & = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\
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| − | & = \sum_{i = 0}^{p - 1} \binom{p}{i} \\
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| − | & = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\
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| − | & = 2^p - 1 .
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| − | \end{align*}
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| − | </cmath>
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| − | For <math>q = 2</math> and <math>p \geq q</math>, we get
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| − | <cmath>
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| − | \begin{align*}
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| − | N \left( p , 2 \right)
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| − | & = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\
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| − | & = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\
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| − | & = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right)
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| − | - \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\
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| − | & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right)
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| − | - p \\
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| − | & = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\
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| − | & = 3^p - 2^p - p .
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| − | \end{align*}
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| − | </cmath>
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| − | For <math>q = 3</math> and <math>p \geq q</math>, we get
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| − | <cmath>
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| − | \begin{align*}
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| − | N \left( p , 3 \right)
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| − | & = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\
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| − | & = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\
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| − | & = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right)
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| − | - \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\
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| − | & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i}
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| − | - \binom{p}{i} \left( p - i \right) \right)
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| − | - \frac{3}{2} p \left( p - 1 \right) \\
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| − | & = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p
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| − | - \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1}
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| − | - \frac{3}{2} p \left( p - 1 \right) \\
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| − | & = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) .
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| − | \end{align*}
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| − | </cmath>
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| − | For <math>q = 4</math> and <math>p = 5</math>, we get
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| − | <cmath>
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| − | \begin{align*}
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| − | N \left( 5 , 4 \right)
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| − | & = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\
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| − | & = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\
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| − | & = \boxed{\textbf{(E) 1296}} .
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| − | \end{align*}
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| − | </cmath>
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| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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| − | ==Solution 2 (Cheese)==
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| − | Let the set of all valid sequences be <math>S</math>.
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| − | Notice that for any sequence <math>\{a_1,a_2,a_3,a_4,a_5\}</math> in <math>S</math>, the sequences
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| − | <cmath>
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| − | \begin{align*}
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| − | \{a_2,a_3,a_4,a_5,a_1\}\\
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| − | \{a_3,a_4,a_5,a_1,a_2\}\\
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| − | \{a_4,a_5,a_1,a_2,a_3\}\\
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| − | \{a_5,a_1,a_2,a_3,a_4\}
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| − | \end{align*}
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| − | </cmath>
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| − | must also belong in <math>S</math>. However, one must consider the edge case all 5 elements are the same (only <math>\{0,0,0,0,0\}</math>), in which case all sequences listed are equivalent. Then <math>\lvert S \rvert \equiv 1 \pmod 5</math>, which yields <math>\boxed{\textbf{(E) 1296}}</math> by inspection.
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| − | ~Tau
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| − | == Video Solution By ThePuzzlr ==
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| − | https://youtu.be/qWIYzgqgCNg
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| − | ~ MathIsChess
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| − | ==Video Solution==
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| − | https://youtu.be/mj78e_LnkX0
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| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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| − | == Video Solution By OmegaLearn using Complementary Counting ==
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| − | https://youtu.be/jWoxFT8hRn8
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| − | ~ pi_is_3.14
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| − | == See Also ==
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| − | {{AMC10 box|year=2022|ab=A|num-b=23|num-a=25}}
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| − | {{AMC12 box|year=2022|ab=A|num-b=23|num-a=25}}
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| − | {{MAA Notice}}
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