Probability HW#1 - Solution
Due: March 18, 2010
Problem 1 (2 points). Seven different gifts are to be distributed among 10
children. How many distinct results are possible if no child is to receive more
than one gift?
Solution. There are
10
× 7! = 604800
7
ways to distribute the gifts.
Problem 2 (2 points). If 12 people are to be divided into 3 committees of
respective sizes 3, 4, and 5, how many divisions are possible?
Solution. There are
12 9 5
= 27720
4 5
3
possible divisions.
Problem 3 (6 points). We have 20 thousand dollars that must be invested
among 4 possible opportunities. Each investment must be integral in units
of 1 thousand dollars, and there are minimal investments that need to be
made if one is to invest in these opportunities. The minimal investments are
2, 2, 3, and 4 thousand dollars. How many different investment strategies
are available if
(a) an investment must be made in each opportunity?
(b) investments must be made in at least 3 of the 4 opportunities.
Solution.
(a) Let xi be the number of investments made in i-th opportunity, si be the
minimal investment needed toPbe made in the i-th opportunity. That is,
s1 = 2, s2 = 2, s3 = 3, s4 = 4, 4i=1 si = 11. Hence we have
x1 + x2 + x3 + x4 = 20,
1
where xi ≥ si , 1 ≤ i ≤ 4. Now defining x′i = xi − si for 1 ≤ i ≤ 4, the
equation becomes
x′1
+
x′2
+
x′3
+
x′4
= 20 −
4
X
si = 20 − 11 = 9,
i=1
where
x′i
≥ 0, 1 ≤ i ≤ 4. The number of different strategies is given by
12
9+4−1
= 220
=
9
4−1
(b) That is, an investment can be made in each opportunity or investments
must be made in 3 of the 4 opportunities. The first case is the problem
in part (a), which has 220 possible strategies. For the second case, we
need to choose 3 of the 4 opportunities. Let i1 , i2 , i3 be the index of the 3
opportunities, and i4 be the index of the rest opportunity. The problem
becomes
xi1 + xi2 + xi3 = 20
where xij ≥ sij for 1 ≤ j ≤ 4. Defining x′ij = xij − sij for 1 ≤ j ≤ 3.
The equation becomes
x′i1 + x′i2 + x′i3 = 20 −
4
X
sij + si4 = 20 −
j=1
4
X
si + si4 = 9 + si4 ,
i=1
where x′ij ≥ 0 for 1 ≤ j ≤ 3. Hence there are
11 + si4
9 + si4 + 3 − 1
=
2
3−1
ways. Then in total we have
4 X
15
14
13
13
11 + si4
= 572
+
+
+
= 220 +
220 +
2
2
2
2
2
i =1
4
choices.
Problem 4 (4 points). Give a combinatorial argument (no computations
are needed) to establist the following identity:
X
n i−1
n
,
=
k−1
k
i=k
where n ≥ k. Hint: Consider the set of numbers 1 through n. How many
subsets of size k have i as their highest-numbered member?
2
Solution. Consider
the set of numbers 1 through n. The number of subsets
of size k is nk . The number of subsets of size k have k as their highest
k−1
numbered member is k−1
, since we need to choose k − 1 elements from 1
to k − 1, and add the element k. Similarly, The number of subset of size k
i−1
have i (i ≥ k) as their highest-numbered member is k−1
. Hence
X
n i−1
n
=
k−1
k
i=k
Problem 5 (5 points). In how many ways can n identical balls be distributed
into r urns so that P
the i-th urn contains at least mi balls, for each i = 1, . . . , r?
Assume that n ≥ ri=1 mi .
Solution. Let xi be the number of balls be distributed into i-th urn, 1 ≤ i ≤ r.
Then we have
x1 + x2 + · · · + xr = n,
where xi ≥ mi . Now defining x′i = xi − mi . Writing the equation as
x′1
+
x′2
+···+
x′r
=n−
r
X
mi ,
i=1
where x′i ≥ 0. So there’s
n−
Pr
mi + r − 1
r−1
i=1
ways to distribute the balls.
Problem 6 (3 points). How many 5-digit numbers can be formed from the
integers 1, 2, . . . , 9 if no digit can appear more than twice? (For instance,
41434 is not allowed.)
Solution. There are 3 cases. First, the 5-digit number is of the form abcde,
we have
9
× 5! = 15120
5
possible numbers. Second, the 5-digit number is of the form aabcd, we have
5!
9 8
· = 30240
1 3
2!
possible numbers. Third, the 5-digit number is of the form aabbc, we have
5!
9 7
·
= 7560
2 1
2!2!
3
possible numbers. Hence, the total number of solutions is 15120 + 30240 +
7560 = 52920.
Problem 7 (4 points). Determine the number of vectors (x1 , . . . , xn ) such
that each xi is a positive integer and
n
X
xi ≤ k,
i=1
where k ≥ n.
Solution. Let x′i = xi − 1 ≥ 0. Then the equation becomes
n
X
x′i
=
i=1
n
X
(xi − 1) ≤ k − n,
i=1
where x′i ≥ 0. Add appropriate t ≥ 0 to make the equality hold. Hence
t+
n
X
x′i = k − n,
i=1
where x′i ≥ 0, t ≥ 0. This gives
k
k − n + (n + 1) − 1
=
n
(n + 1) − 1
ways.
Problem 8 (6 points). If there are no restrictions on where the digits and
letters are placed, how many 8-place license plates consisting of 5 letters and
3 digits are possible if no repetitions of letters or digits are allowed? What
if the 3 digits must be consecutive?
Solution. If no repetitions of letters or digits are allowed, we have
10 26
· 8! = 318269952000
5
3
ways. If the 3 digits must be consecutive, we have
10 26
· 6! · 3! = 34100352000
5
3
ways.
4