AMATYC SML Training

For more problems, please visit the AMATYC website at www.amatyc.org

For solutions to last week’s problems, visit my website at www.montgomerycollege.edu/~rpenn

 

1.  How many different arrangements of the letters of AMATYC are possible?

A.  120                        B.  240                        C.  360                        D.  540                        E.  720

 

There are 6!=720 ways to arrange 6 symbols, but this double counts all the arrangements of these letters, as the 2 A’s can be swapped and the result would look the same.  So the answer is C, 260.

 

2.  Consider all arrangements of the letters of AMATYC with either the A’s together or the A’s on the ends.  What fraction of all possible arrangements satisfies these conditions?

A.  1/5                         B.  2/15                       C. 1/3              D. 2/5              E. 3/5

 

There are 360possible arrangements of the letters (as shown in #1). 

There are 6 ways the A’s can be placed (spots 1&2, 2&3, 3&4, 4&5, 5&6 or 6&1), and for each there are 4! ways the other 4 letters can be arranged, for a total of 6*4!=144.  So the fraction is 144/360 = 2/5, so D is the answer.

 

3.  Let A= {0,1,2,3,4,5,6,7,8,9}.  How many 3-element subsets of A contain at least two consecutive integers?

A.  32              B.  40              C.  48              D.  56              E.  64

 

If A contains 0 and 1, there are 8 other options for the 3rd element, 2-9.

If the smallest consecutive integer elements are 1 and 2, there are 7 other options for the 3rd element, 3-9.  Similarly for each of the other 7 possible consecutive integer elements there are 7 ways to complete the subset.  So there are 8 + 8*7 = 64 such subsets.

 

4. In how many ways can slashes be placed among the letters AMATYCSML to separate them into four groups with each group containing at least one letter?

A.  28              B.  56              C. 70               D. 84               E. 112

 

Imagine there are spaces where slashes can be placed between the letters.  There are 8 such spaces.  We need to choose 3 of them to contain slashes, so there are C(8,3) = 56 such ways.

 

 

5.  In how many ways can 9 identical dominos (2x1 rectangles) be used to exactly cover a 3x6 rectangle with no overlap?  Assume 2 coverings are different if the 9 dominos are not in exactly the same positions.

A. 21               B. 31               C. 35               D. 41               E. 47

First note that there are exactly 3 ways that 3 dominos can cover 2 columns:


                                   

 

 

 

 

 

 


So if the dominos are placed in such a way as to have 3 cover the 1st 2 columns, 3 cover the next 2, and 3 cover the last 3, that can be done in 3*3*3 = 27 ways.

Now, how many ways can the dominos be placed to cover 4 columns (excluding cases that would have already been counted?

This would require a horizontal domino across columns 2 and 3, and that in turn can only be done in 2 ways:  as shown, or the upside down version of this.

 


                                               

For each of these, there are 3 ways to fill the last 2 columns

                                                so there are 2*3 = 6 possibilities.  There are similarly 6       

                                                possibilities that fill the first 2 columns evenly, then have a

                                                pattern such as this in columns 3-6.  So we are now up to

                                                39 total ways.   All that remains is finding ways that cover all 6 columns without any “clean breaks” before the end.  There are 2 of these:

 

 

 

 

 

 

 

 


6.  A sock drawer contains 6 identical black socks, and 4 identical brown socks.  If you reach in and randomly select 2 socks, what is the probability that they will be a matched pair?

A.  2/5                         B. 7/15                                    C. ½                D.  8/15           E.  3/5

 

(C(6,2) +C(4,2) ) / C(10,2) = 21/45 = 7/15

7.  A positive integer less than 1000 is chosen at random.  What is the probability it is a multiple of 3, but a multiple of neither 2 nor 9?

A.  1/10                       B.  1/9                         C. 1/8              D. 2/9              E. 1/3

 

There are 999/3 = 333 numbers less than 1000 that are divisible by 3.

There are 999/9 = 111 numbers less than 1000 that are divisible by 9, all of which must also be divisible by 3 (since 9 itself is a multiple of 3).

A number being divisible by both 2 and 3 is equivalent to it being divisible by 6.  There are 166 such numbers less than 1000 (999/6, rounded down).

 

If we take the 333 numbers divisible by 3, and subtract out the 111 that are divisible by 9 and the 166 that are divisible by 6, that would seem to give the count of integers less than 1000 divisible by 3 but not 2 or 9.  However, some numbers are divisible by both 6 and 9, and so are “removed” from the count twice!  We need to figure out how many of these numbers there are, and add that number to compensate for the double subtraction.

A number that being divisible by both 6 and 9 is equivalent to it being divisible by 18, and 999/18 55 (rounded down again).   So there are 333-111-166+55 = 111 of the 999 integers that meet the requirement, and the probability is 111/999, or 1/9.

 

 

8.  Teams A and B play a series of games; whoever wins the two games first wins the series.  If Team A has a 70% chance of winning any single game, what is the probability that Team A wins the series?

A.  .616                       B.  .637                       C. .657                        D. .700                        E.  .784

 

A can win the first game and then the 2nd, with probability .7*.7 = .49.

A can win, then lose, then win again, with probability .7*.3*.7 = .147

A can lose, then win, then win, with probability .3*.7*.7. 

Adding these results gives the probability of A winning the series, .784.

 

 

9.  A bag holds 5 cards identical except for color. Two are red on both sides, two are black on both sides, and one is red on one side and black on the other.  If you pick a card at random and the only side you can see is red, what is the probability that the other side is also red?

A.  ½               B. 2/3              C.  ¾               D. 4/5              E.  5/6

 

There are 5 sides you could be looking a that are red (either side of the 2 red cards, or the red side of the 2-colored card).  On 4 of those, if you turn it over you will see red, so 4/5.