2015 Mathcounts National sprint #22 level 1 +

 

2015 Mathcounts National sprint #22

If six people randomly sit down at a table with six chairs, and they do not notice that there are name tags marking assigned seats, what is the probability that exactly three of them sit in the seat he or she was assigned? Express your answer as a common fraction.

Try this yourself first (extremely), then scroll down for solutions.

Method 1-Elementary Counting

Choose the three people who sit correctly: $(\genfrac{}{}{0ex}{}{6}{3})=20$.
The remaining three must all sit incorrectly. A quick check (or listing) shows only two ways: $(ABC)$ or $(ACB)$.
Total favourable seatings $=20\times 2=40$; total seatings $=6!$.
Therefore $P={\displaystyle \frac{40}{720}}=\boxed{{\displaystyle \frac{1}{18}}}$.

Method 2 – Derangement formula

Pick the three fixed seats: $(\genfrac{}{}{0ex}{}{6}{3})=20$. Derange the other three: $!3=2$.
Again $20\times 2=40$ good seatings, so $$P=\frac{20\cdot !3}{6!}=\boxed{{\displaystyle \frac{1}{18}}}.$$

Derange the remaining 3 people. The number of derangements of 3 items is $$D_3=3!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!})=6(1-1+\frac{1}{2}-\frac{1}{6})=2.$$

student learning log, a high school student I volunteer

each week a space 

SAT 14 #2 wrong , 15 #2 

SAT 16, #5, 6, wrong , 17, 3 and 6 

2 weeks later  

18, #5 and 5

19  vocab. words 11 - 20 

SAT practice test 8, later math don't know what to do 

2 weeks later 

20, #5, 21, #4

more than a month later 

22, #4 and 5 , 23, #4, 6, 8, 12

30, #5, 31, #9 

32, #4, 5, 33, 2, 9 

local math competition : 21-22  meet 1, # 1, 2, 3 right 

34, #4, 35, #6 and 8 

21-22 math competition meet 2, #1 and 3 , #2 almost 

36 #1，2，3  need to work on harder vocabulary words

37 # 6，9

Math 2021-22 meet 2 # 1 and 3 

5/6/25 This week for reading I did Crack Sat tests 38 and 39 and I only gave myself 15 minutes for each test to time myself. 

For test 38 the questions I got wrong were 1 and 11 and for test 39 the questions I got wrong were 2 and 5. 

Although I felt rushed I was able to complete the test while still comprehending everything so that’s good. 

For math this week I realized that I struggled a lot on the algebra 2 part of the (local math competition) tests so I decided that I should learn some of the curriculum. 

To learn the course I went on Khan academy and did 2 units of the course and I plan on continuing learning the course to help with (local math competition) problems in the future. Thank you.

AOPS videos

6/2, 2025 one interesting question that has nice solution : Beginning level

 

2010 Mathcounts Nationals sprint :

22. Side AB of regular hexagon ABCDEF is extended past B to point X such that AX = 3AB. Given that each side of the hexagon is 2 units long, what is the length of segment FX? Express your answer in simplest radical form.

Try this question first before you scroll down for the solution. 

Then when we draw line FX, and by the Pythagorean Theorem, we have $$FX=\sqrt{(\sqrt{3}{)}^2+(1+2+4{)}^2}=\sqrt{3+49}=\boxed{{\displaystyle 2\sqrt{13}}}.$$