Sunday, October 29, 2017

On Many Different Cases to Consider

Problem 1 Part b – Correct! – Score: 3 / 7 (1923)
We call a number special if every digit in the number either is a 1 or borders a 1. For example, 11111, 13, 141, 1441, 515151, and 101 are all special, but 10001, 222, 122, and 1333 are not special.

How many positive 3-digit numbers are special?
Solution:
A 3-digit number is special if it has a 1 in the middle, or it both starts and ends in a 1. There are $9\cdot 1\cdot 10$ of the first type (we have 9 choices for the first digit and 10 choices for the last digit). There are also 10 of the second type (we must choose the middle digit), but we already counted 111 in the first case, so we don't count it again; we get 9 new special numbers in the second case. Therefore, there are 90 + 9 = 99 special three-digit numbers.
Your Response(s):
  • :( 81
  • :( 82
  • :) 99

Analysis: The formats 1_1, _1_, and 111 aren't enough - the numbers can be in the form of 11_ or _11 too!

Saturday, October 28, 2017

On Casework and Different Things

Solution

In topic: Distinguishability (Counting & Probability).
 
Correct! Way to go!
Matt's four cousins are coming to visit. There are four identical rooms that they can stay in. If any number of the cousins can stay in one room, how many different ways are there to put the cousins in the rooms?
Your First Answer: 14
Your Second Answer: 15
Solution:
Just counting the number of cousins staying in each room, there are the following possibilities: (4,0,0,0), (3,1,0,0), (2,2,0,0), (2,1,1,0), (1,1,1,1).

(4,0,0,0): There is only $1$ way to put all the cousins in the same room (since the rooms are identical).

(3,1,0,0): There are $4$ ways to choose which cousin will be in a different room than the others.

(2,2,0,0): Let us consider one of the cousins in one of the rooms. There are $3$ ways to choose which of the other cousins will also stay in that room, and then the other two are automatically in the other room.

(2,1,1,0): There are $\binom{4}{2}=6$ ways to choose which cousins stay the same room.

(1,1,1,1): There is one way for all the cousins to each stay in a different room.

The total number of possible arrangements is $1+4+3+6+1=\boxed{15}$.

Symmetry, Reflection, and When to Divide

  Level 25 Counting & Probability  +123 XP

Review

In topic: Counting with Symmetry (Counting & Probability).
 
Incorrect. Oops!
In how many ways can $7$ people sit around a round table if no two of the $3$ people Pierre, Rosa, and Thomas can sit next to each other?
Your First Answer: 384
Your Second Answer: 1008
Solution:
After Pierre sits, we can place Rosa either two seats from Pierre (that is, with one seat between them) or three seats from Pierre. We tackle these two cases separately:

Case 1: Rosa is two seats from Pierre. There are $2$ such seats. For either of these, there are then four empty seats in a row, and one empty seat between Rosa and Pierre. Thomas can sit in either of the middle two of the four empty seats in a row. So, there are $2\cdot 2 = 4$ ways to seat Rosa and Thomas in this case. There are then $4$ seats left, which the others can take in $4! = 24$ ways. So, there are $4\cdot 24 = 96$seatings in this case.

Case 2: Rosa is three seats from Pierre (that is, there are $2$ seats between them). There are $2$ such seats. Thomas can't sit in either of the $2$seats directly between them, but after Rosa sits, there are $3$ empty seats in a row still, and Thomas can only sit in the middle seat of these three. Once again, there are $4$ empty seats remaining, and the $4$ remaining people can sit in them in $4! = 24$ ways. So, we have $2\cdot 24 = 48$seatings in this case.

Putting our two cases together gives a total of $96+48 = \boxed{144}$ seatings.

Distinguishability is Annoying

  Level 25 Counting & Probability  +125 XP

Review

In topic: Distinguishability (Counting & Probability).
Source: AoPS Staff
Incorrect. Oops!
Rachel has two identical basil plants and an aloe plant. She also has two identical white lamps and two identical red lamps she can put each plant under (she can put more than one plant under a lamp, but each plant is under exactly one lamp). How many ways are there for Rachel to put her plants under her lamps?
Your First Answer: 18
Your Second Answer: 12
Solution:
We can split this up into cases.

First, consider the case when all three plants are under the same color lamp. Either all three plants are under the same lamp, both basil plants are under one lamp and the aloe plant is under the other lamp, or the aloe plant and one basil plant are under one lamp and the other basil plant is under the other lamp. This case gives us three possibilities for each color of lamp, for a total of six possibilities.

Next, consider the case where the aloe plant is under a different color of lamp than the two basil plants. Since the two lamps of the same color the aloe plant can be under are identical, it doesn't matter which one the aloe plant is under. The basil plants can either both be under the same lamp, or each be under a different lamp. This case gives us two possibilities when the aloe is under a white lamp and two possibilities when the aloe is under a red lamp, for a total of four possibilities.

Last, consider the case where the basil plants are each under a different colored lamp. The aloe plant can be under the same white lamp as a basil plant, the same red lamp as a basil plant, a different white lamp from the basil plant, or a different red lamp from the basil plant, for a total of four possibilities. In all, there are $6+4+4=\boxed{14}$ possibilities.

Is that Divisible?

Question text A number 32 a 5 b ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ is divisible by 11 . Its last two digits form a number 5 b ¯ ¯ ¯ ¯ ¯ tha...