JUNG PAN GYUN
Assignment hw09 due 05/19/2025 at 11:59pm KST
2025-1 Discrete Math GEDB007 45
Problem 1. (1 point)
Find the number of ways to order the letters SUNGKYUNKWAN
with given condition.
Problem 4. (1 point)
How many terms are there in the following expression?
(x + y)20 (u + x + y)(v + x + y)
1. no conditions.
2. no two consecutive K’s.
3. no two consecutive N’s.
A solution to this problem will be available after the due date.
A solution to this problem will be available after the due date.
Answer(s) submitted:
• no response
Answer(s) submitted:
• no response
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• no response
submitted: (incorrect)
recorded: (incorrect)
Problem 5. (1 point)
Compute the coefficient of x3 y4 in the following expansion.
submitted: (incorrect)
recorded: (incorrect)
√
√
√
( x + xy + y)2 (x + y)6
Problem 2. (1 point)
How many different ways are there to give 35 identical candies to
6 children so that every child gets at least 2 candies?
A solution to this problem will be available after the due date.
Answer(s) submitted:
• no response
submitted: (incorrect)
recorded: (incorrect)
Problem 6. (1 point)
Compute the following sums.
A solution to this problem will be available after the due date.
Answer(s) submitted:
• no response
submitted: (incorrect)
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10 ∑
i=0
10
=
i
10
∑ (−1)i
Problem 3. (1 point)
How many integers between 1 and 1,000,000 have the sum of digits equal to 19?
i=0
10
∑ 5i
i=0
10
=
i
(Here,
A solution to this problem will be available after the due date.
10
=
i
n
k means the binomial coefficient C(n, k).)
A solution to this problem will be available after the due date.
Answer(s) submitted:
• no response
Answer(s) submitted:
• no response
• no response
• no response
submitted: (incorrect)
recorded: (incorrect)
submitted: (incorrect)
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1
Problem 7. (1 point)
Compute the following sums.
Problem 9. (1 point)
Order all of the following sentences so that they form a proof for
the following problem:
10
10
∑i i =
i=0
Let X be a subset of {1, 2, . . . , 2n} with n + 1 elements. Show that
there are two distinct elements a, b ∈ X such that a divides b.
10
11 10
=
∑
i
i=0 i + 1
0. Every positive integer k has a unique expression k = 2 j m,
where m is odd.
1. Since X has more than n elements, by the pigeonhole
principle, there are two integers k, k′ ∈ X such that k =
2i m and k = 2 j m with the same odd integer m.
2. If k ∈ {1, 2, . . . , 2n}, then in this expression of k the odd
integer m is one of the n elements in {1, 3, . . . , 2n − 1}.
3. Therefore we always have two distinct elements a, b ∈ X
such that a divides b.
4. Similarly, if j < i, then k′ divides k.
5. If i < j, then k = 2i m divides k′ = 2 j m = 2 j−i · 2i m.
A solution to this problem will be available after the due date.
Answer(s) submitted:
• no response
• no response
submitted: (incorrect)
recorded: (incorrect)
Answer(s) submitted:
• no response
Problem 8. (1 point)
Order all of the following sentences so that they form a proof for
the following problem:
submitted: (incorrect)
recorded: (incorrect)
Problem 10. (1 point)
Order all of the following sentences so that they form a proof for
the following problem:
Suppose that Federer played tennis 198 days among the last 365
days. Show that Federer played tennis on certain two days which
were 30 days apart. (In other words, show that he played tennis
on the ith day and on the jth day for some 1 ≤ i < j ≤ 365 with
j − i = 30.)
Alice drinks at least one cup of coffee every day. She does not
drink more than 12 cups of coffee per week. Show that for the
next 11 weeks, there will exist consecutive days during which she
drinks exactly 21 cups of coffee.
0. Since 31 ≤ b1 < b2 < · · · < b198 ≤ 365 + 30 = 395, the
198 · 2 = 396 integers in a1 , . . . , a198 , b1 , . . . , b198 are contained in {1, 2, . . . , 395}.
1. Then a j − ai = 30, which means that Federer played on
day ai and day a j which are 30 days apart.
2. Let a1 , . . . , a198 be the days that Federer played tennis,
where 1 ≤ a1 < a2 < · · · < a198 ≤ 365.
3. Let’s consider another sequence b1 , . . . , b198 , where bi =
ai + 30.
4. Since ai ̸= a j and bi ̸= b j for all i ̸= j and, we must have
ai = b j = a j + 30 for some 1 ≤ i < j ≤ 198.
5. By the pigeonhole principle, at least two of these 396
numbers are equal.
0. Since 1 ≤ b1 < · · · < b77 = a1 + · · · + a77 = 132, each
number in the sequence S is at least 1 and at most
132 + 21 = 153.
1. Since bi ’s are all distinct and bi + 21’s are all distinct, we
must have bi = b j + 21 for some i and j.
2. For 1 ≤ i ≤ 77, let ai be the number of cups of coffee Alice drinks on the ith day. Then by assumption
a1 + · · · + a77 ≤ 11 · 12 = 132.
3. Let bi = a1 + · · · + ai for 1 ≤ i ≤ 77 and consider the numbers in the sequence
S = (b1 , . . . , b77 , b1 + 21, b2 + 21, . . . , b77 + 21).
Answer(s) submitted:
• no response
4. Since there are 77 · 2 = 154 numbers in S, by the pigeonhole principle, there two equal numbers in S.
5. Then 21 = bi − b j = a j+1 + a j+2 + · · · + ai and Alice
drinks exactly 21 cups of coffee during the consecutive
days from day j + 1 to day i.
submitted: (incorrect)
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Answer(s) submitted:
• no response
2
submitted: (incorrect)
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