1. How many three digit numbers can be formed from the digits 1,2,3,4,5 and 6, if each
digit can only be used once? How many of these are odd numbers? How many are
greater than 330?
There are 6 digits in total. The numbers with 3 digits, with all digits distinct from each
other, are the permutations of the 6 digits taken 3 at a time, and therefore there are 6*5*4
= 120 of them.
To be odd, one such number must end with 1, 3, or 5. We can construct all of the odd
three digit numbers. For example, the odd numbers ending in 1 are the permutation of the
remaining five digits taken two at a time. So the total number of odd numbers is:
3*(5*4)=60.
A number is larger than 330 if its first digit is 4, 5 or 6, or if its first digit is 3 and its
second digit is 4,5 or 6. In the first case, the number of possibilities is 3 times the number
of permutations of the remaining 5 digits taken 2 at a time, i.e. 3*(5*4)=60. In the second
case, the number of possibilities is 3 times the number of permutations of the remaining 4
digits taken 1 at a time, that is 3*4=12. In total, 72.
2. The set S consists of the 10 integers 1 to 10. Find the number of subsets that contain
the integer 1 and/or the integer 3 and/or the integer 7.
Let’s call A the set of subsets of the integers 1 to 10. Call B the set of subsets that contain
the integer 1 and/or the integer 3 and/or the integer 7. Call C the set of subsets that do not
contain 1,3 or 7. Then B and C are disjoint, and A = B∪C. Hence, the cardinality of B
(which is what we are looking for) is equal to the cardinality of A minus the cardinality
of C. Remember that the number of subsets of a set with N elements is 2N. Thus, the
solution is 2 10 - 2 7 = 896.
3. How many possible events are there in a sample space with N possible outcomes?
Given the outcome of a trial, how many events have occurred?
Any subset of the sample space is an event. Hence, there are 2N possible events.
An event occurred when an outcome it contains occurred. The number of events
containing a given outcome is 2N-1. (You obtain this by construction. Any event
containing the outcome x can be constructed as a subset of S/x, the sample space minus x,
united to x. S/x has N-1 elements, so it has 2N-1 subsets.)
4. Find how many different possibilities there are for each poker combination (in a 52
cards deck):
• Full house (three card of the same rank, accompanied by two of the same rank);
• Three of a kind (three cards of the same rank accompanied by two kickers of
different ranks);
• Two pair (two cards of the same rank, two cards of another rank, and one kicker
of a third rank);
• Pair (two cards of the same rank accompanied by three kickers of different ranks);
Full house: For a given rank for the first three cards, there are C43= 4 choices for the first
three cards. For a given rank of the remaining two cards, there are C42= 6 choices. Note
that, given the rank of the first three cards, there are only 12 possible ranks for the second
two cards. Hence, the overall number of possibilities is 13*12*4*6 = 3744.
Three of a kind: For a given rank for the first three cards, there are C43= 4 choices for the
suite distribution of such cards. As for the kickers, we can choose two out of the
remaining 49 cards, such that they have different rank from each other (otherwise it is a
full house) and that both have rank different from the first three (otherwise it is a four of a
kind). Let’s first choose the ranks for these two cards. There are 12 ranks left, and
therefore there are C122 combinations of 12 ranks taken 2 at a time. As for the suite, for
each such rank combination there are 4 possible choices for the first card and 4 possible
choices for the second card. Overall, 13*C43* C122*4*4 = 54912.
You could also reason in this way. Among the remaining 49 cards, there are 12*C42 pairs
that have identical rank, and 48 pairs such that one of them has rank equal to the first
three cards. Thus, there are C492- 12*C42 –48 = 1056 choices for the second two cards.
Overall, 4*13*1056 = 54912 possibilities.
Two pair: Let’s construct all the possible hands by starting with the kicker (13*4
possibilities). Among the remaining 12 ranks, let’s choose two different ranks for the two
pairs (C122 possibilities). Each pair has C42 choices for the suites. Overall, 13*4* C122*
C42* C42 = 123552 possibilities.
Pair: For a given rank for the first two cards (13 choices) , there are C42= 6 choices for
the suite distribution. Let’s choose 3 different ranks from the 12 remaining (C123
combinations). Each such combination accepts 4*4*4 possibilities for the suite
distribution. Overall, 13*C42*C123*4*4*4 = 1098240 combinations.
5. Consider a computer that uses binary words with 8 bits.
• How many words are there with 4 or less bits equal to “0”?
• How many words are there with 6 or more bits equal to “0”?
The number of words with n bits equal to “0” can be obtained in this way. Consider the
indices of the bits in a word (1,2,…,8). We extract n of these indices, and assign “0” to
their bits, and “1” to the other bits. Thus, there are C8n such words. To find the number of
words that have ≤ 4 bits equal to “0”, we need to consider the set of words with 0 bits
equal to “0”, the set of words with 1 bit equal to “0”, and so forth till we exhaust the
possibilities. These sets are disjoint, so we can just sum their cardinalities:
4 " %
8
($# i '& = 1+ 8 + 28 + 56 + 70 = 163 . The number of words with 6 or more bits equal to “0”
i= 0
8 " %
8
is: ($ ' = 28 + 8 + 1 = 37 .
i
i= 6 # &
!
!
6. You want to carry 6 identical objects using 3 cases. One case contains exactly 3
objects, one case contains exactly 2 objects, and one case contains exactly 1 object.
How many different ways there are to distribute the objects in the cases? (Assume
that the order of the objects in the same case is irrelevant.)
You first place 3 objects in the first case (C63 possibilities), then 2 of the remaining 3
objects in the second case (C32 possibilities), and finally you place the remaining object in
the last case. Overall, C63* C32= 20*3=60.
7. Starting from the letters “C E 1 0 7”, how many different words with ≤ 3 letters can
you build (without ever reusing a letter in the same word)?
The set of words with n letters is the set of permutation of 5 letters taken n at a time
(order matters here!): P5n. The number of different words with ≤ 3 letters is
3
therefore " Pn5 = 5 + 20 + 60 = 85 .
n=1
8. Give a formal proof (not based on Venn diagrams) of the following identities:
!•
( M " B ) " ( M # B) = $
• ( M " B ) # ( M # B) = S
!
!
Based on the previous identities, prove that M ! B = M " B
•
•
( M ! B ) ! ( M " B ) = ( M ! B ! M) " ( M ! B ! B ) = # ! # = #
( M ! B ) " ( M " B ) = ( M " M " B) ! ( B " M " B ) = S ! S = S
Then, note that, if U ! V = " and U ! V = S , then V = U . Thus, M ! B = M " B