A Skill for the 21st Century: Problem Solving by Richard Rusczyk, founder of "Art of Problem Solving".

Top 10 Skills We Wish Were Taught at School, But Usually Aren't

from Lifehacker

## Wednesday, July 1, 2015

## Wednesday, June 24, 2015

### Problem Solving Strategy: Complementary Counting

**#: How many three digit numbers contain the digit "9" at least once?**

**Solution:**

For the "at least" questions: a lot of the time, the easiest way to solve the problems is to use the total number of ways minus the number of ways which do not satisfy the criteria we're looking for.

There are 999 -100 + 1 = 900 three digit numbers.

There are 8 x 9 x 9 = 648 numbers that do not contain the digit "9" at all, so the answer is 900 - 648 =

**252**

**#2: How many three digit numbers have at least two digits that are the same?**

Solution:

Use complementary counting to find how many three digit numbers have no digit that are the same.

9 digit choices for the hundredth digit (no 0), 9 for the tenth digit (0 is allowed) and 8 digit left for the unit digit, so 9 * 9 * 8 = 648

900 (3 digit numbers) - 648 =

**252**

**#3: How many of the natural numbers from 1 to 600, inclusive, contain the digit "5" at least once?**

**(The numbers 152 and 553 are two natural numbers that contain the digit 5 at least once, but 430 is not.)**

Solution:

From 1 to 600 inclusive, there are 600 - 1 + 1 = 600 numbers.

___ ___ ___ If we're looking for all the three-digit numbers which do not contain any 5s: In the hundreds place, you can put 0, 1, 2, 3, or 4 (we'll get rid of '000' later). In the tens place, you can put 9 digits, and there are another 9 possible digits for the units digit, so 5 x 9 x 9 = 405.

405 - 1 (to get rid of '000') + 1 (to compensate for not counting the number '600') = 405, which is the total number of numbers which don't use the digit '5'.

600 - 405 =

**195**, which is the answer.

**#4: If you toss three coins, what is the probability that at least one coin lands heads up?**

Solution:

There is a \(\dfrac {1} {2}\times \dfrac {1} {2}\times \dfrac {1} {2}=\dfrac {1} {8}\)chance to get all tails = no heads, so if we want at least one head, the answer is 1 - \(\dfrac {1} {8}\)= \(\dfrac {7} {8}\).

**Other applicable problems (answers below):**

**#1: Amy tosses a nickel four times. What is the probability that she gets at least as many heads as tails ?**

**#2: What is the probability that the product of the top faces on 2 standard die is even when rolled?**

**#3: 3 numbers are selected at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 3 numbers is even ? Express your answer as a common fraction in lowest terms.**

**#4: 9 fair coins are flipped. What is the probability that at least 4 are heads?**

**#5: How many 3 digit numbers does not contain the digit 1 but have at least one digit that is 5?**

Answer: #1-\(\dfrac {11} {16}\) ; #2-\(\dfrac {3} {4}\) ; #3- \(\dfrac {7} {8}\) ; #4-\(\dfrac {191} {256}\) #5:

**217**

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