Showing posts with label casework. Show all posts
Showing posts with label casework. Show all posts

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 $1+4+3+6+1=\boxed{15}$.

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...