|
|
| (32 intermediate revisions by 7 users not shown) |
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #redirect [[2022 AMC 10A Problems/Problem 24]] |
| − | | |
| − | 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
| |
| − | 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
| |
| − | 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
| |
| − | does not contain at least <math>2</math> digits less than <math>2</math>.)
| |
| − | | |
| − | <math>\textbf{(A)} \, 500 \qquad\textbf{(B)} \, 625 \qquad\textbf{(C)} \, 1089 \qquad\textbf{(D)} \, 1199 \qquad\textbf{(E)} \, 1296 </math>
| |
| − | | |
| − | ==Solution==
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | We have the following recursive equation:
| |
| − | <cmath>
| |
| − | N \left( p, q \right)
| |
| − | = \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1
| |
| − | </cmath>
| |
| − | and the boundary condition <math>N \left( p, 0 \right) = 1</math> for any <math>p \geq 0</math>.
| |
| − | | |
| − | By solving this recursive equation, for <math>q = 1</math> and <math>p \geq q</math>, we get
| |
| − | <cmath>
| |
| − | \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*}
| |
| − | </cmath>
| |
| − | | |
| − | For <math>q = 2</math> and <math>p \geq q</math>, we get
| |
| − | <cmath>
| |
| − | \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*}
| |
| − | </cmath>
| |
| − | | |
| − | For <math>q = 3</math> and <math>p \geq q</math>, we get
| |
| − | <cmath>
| |
| − | \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*}
| |
| − | </cmath>
| |
| − | | |
| − | For <math>q = 4</math> and <math>p = 5</math>, we get
| |
| − | <cmath>
| |
| − | \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*}
| |
| − | </cmath>
| |
| − | | |
| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| |
| − | | |
| − | ==Solution 2 (Cheese)==
| |
| − | Let the set of all valid sequences be <math>S</math>.
| |
| − | Notice that for any sequence <math>\{a,b,c,d,e\}</math> in <math>S</math>, the sequences
| |
| − | <cmath>
| |
| − | \begin{align*}
| |
| − | \{b,c,d,e,a\}\\
| |
| − | \{c,d,e,a,b\}\\
| |
| − | \{d,e,a,b,c\}\\
| |
| − | \{e,a,b,c,d\}
| |
| − | \end{align*}
| |
| − | </cmath>
| |
| − | must also belong in <math>S</math>. The only exception is when all 5 elements are the same, i.e. <math>\{0,0,0,0,0\}</math>. Then <math>\lvert S \rvert \equiv 1 \pmod 5</math>, which by inspection of the answer choices yields <math>\boxed{(E) 1296}</math>.
| |
| − | | |
| − | ~Tau
| |
| − | | |
| − | == Video Solution By ThePuzzlr ==
| |
| − | https://youtu.be/qWIYzgqgCNg
| |
| − | | |
| − | ~ MathIsChess
| |
| − | | |
| − | ==Video Solution==
| |
| − | | |
| − | https://youtu.be/mj78e_LnkX0
| |
| − | | |
| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| |
| − | | |
| − | == Video Solution By OmegaLearn using Complementary Counting ==
| |
| − | | |
| − | https://youtu.be/jWoxFT8hRn8
| |
| − | | |
| − | ~ pi_is_3.14
| |
| − | | |
| − | == See Also ==
| |
| − | | |
| − | {{AMC12 box|year=2022|ab=A|num-b=23|num-a=25}}
| |
| − | {{AMC10 box|year=2022|ab=A|num-b=23|num-a=25}}
| |
| − | {{MAA Notice}}
| |
