2003 AMC 12B Problems/Problem 19

Problem

Let $S$ be the set of permutations of the sequence $1,2,3,4,5$ for which the first term is not $1$ . A permutation is chosen randomly from $S$ . The probability that the second term is $2$ , in lowest terms, is $a/b$ . What is $a+b$ ?

$\mathrm{(A)}\ 5 \qquad\mathrm{(B)}\ 6 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 16 \qquad\mathrm{(E)}\ 19$

Solution

There are $4$ choices for the first element of $S$ , and for each of these choices there are $4!$ ways to arrange the remaining elements. If the second element must be $2$ , then there are only $3$ choices for the first element and $3!$ ways to arrange the remaining elements. Hence the answer is $\frac{3 \cdot 3!}{4 \cdot 4!} = \frac {18}{96} = \frac{3}{16}$ , and $a+b=19 \Rightarrow \mathrm{(E)}$ .

Solution 2

There is a $\frac {1}{4}$ chance that the number $1$ is the second term. Let $x$ be the chance that $2$ will be the second term. Since $3, 4,$ and $5$ are in similar situations as $2$ , this becomes $\frac {1}{4} + 4x = 1$

Solving for $x$ , we find it equals $\frac {3}{16}$ , therefore $3 + 16 = 19 \Rightarrow \mathrm{(E)}$

Solution 3

Let's focus on 2, 3, 4, and 5 right now. There are $4!$ ways to arrange these numbers. We can "insert" 1 into the arrangement after the first, second, third, and fourth numbers. There are $4! \cdot 4 = 96$ ways to do this.

In 24 of those ways, 1 is in the second position, and in the remaining 72 sequences, 2, 3, 4, and 5 occur in the second position the same number of times. $\frac{\frac{72}{4}}{96} = \frac{3}{16}$ , therefore $3 + 16 = 19 \Rightarrow \mathrm{(E)}$

Solution 4

The probability could be written as a fraction where $\frac {\text{number of viable solutions in S}}{\text{total permutations in S}}$ .

The total number of permutations is essentially $5!$ , and the number of permutations where $1$ is the first number is $4!$ , therefore the number of permutations in $S$ is $5!-4!$ .

For the first number in the desired solution, there are 3 options ( $3$ , $4$ , $5$ ). For the second number in the desired solution, there is only one option ( $2$ ). For the third number in the desired solution, there are 3 options ( $1$ , and the numbers not used in the first digit). For the fourth number in the desired solution, there are 2 options (numbers not used in the third digit). For the last number in the desired solution, there is only one option (number not used in the fourth digit).

Therefore, $3 \cdot 1 \cdot 3 \cdot 2 \cdot 1$ will be the number of desired outcomes.

Finally, $\frac {3 \cdot 1 \cdot 3 \cdot 2 \cdot 1}{5!-4!}$ equates to $\frac {18}{96}$ , which is $\frac {3}{16}$ .

The answer is then $3+16=\boxed{\textbf{(E)}\ 19}$ .

~ Tyrone12345

Video Solution by OmegaLearn

https://youtu.be/IRyWOZQMTV8?t=1215

~ pi_is_3.14

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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