MATH 140 - Lecture 8 Worksheet 1. (a) How many lists of length 3 are there, if the first two entries are letters from the English alphabet, and the last entry is a number? (b) How many lists of length 4 can be made from the letters A,B,C,D, if repetitions are allowed? (c) How many lists of length 4 can be made from the letters A,B,C,D, if repetitions are not allowed? (d) How many lists of length 3 can be made from the lottora RCD :C
MATH 140 - Lecture 8 Worksheet 1. (a) How many lists of length 3 are there, if the first two entries are letters from the English alphabet, and the last entry is a number? (b) How many lists of length 4 can be made from the letters A,B,C,D, if repetitions are allowed? (c) How many lists of length 4 can be made from the letters A,B,C,D, if repetitions are not allowed? (d) How many lists of length 3 can be made from the lottora RCD :C
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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How can we complete questions 1a-1e?
![**MATH 140 - Lecture 8 Worksheet**
1.
(a) How many lists of length 3 are there, if the first two entries are letters from the English alphabet, and the last entry is a number?
(b) How many lists of length 4 can be made from the letters A, B, C, D, if repetitions are allowed?
(c) How many lists of length 4 can be made from the letters A, B, C, D, if repetitions are not allowed?
(d) How many lists of length 3 can be made from the letters A, B, C, D, if repetitions are not allowed and each list must contain "C"?
(e) A byte is an 8-digit binary string, which is a list of 1’s and 0’s. How many such byte strings are there? How many end in 0? How many have the second and fourth digit as 1’s?
2.
(a) Compute 3!, 4!, 5!, 10!.
(b) How many 9-digit numbers (using 1, 2, 3, ..., 9) are there if repetition is not allowed? How many are there which have at least one digit repeated?
(c) Compute \(\frac{100!}{98!}\).
The calculation shown is:
\[
= \frac{100 \cdot 99 \cdot 98!}{98!} = 100 \cdot 99 = 9900
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Transcribed Image Text:**MATH 140 - Lecture 8 Worksheet**
1.
(a) How many lists of length 3 are there, if the first two entries are letters from the English alphabet, and the last entry is a number?
(b) How many lists of length 4 can be made from the letters A, B, C, D, if repetitions are allowed?
(c) How many lists of length 4 can be made from the letters A, B, C, D, if repetitions are not allowed?
(d) How many lists of length 3 can be made from the letters A, B, C, D, if repetitions are not allowed and each list must contain "C"?
(e) A byte is an 8-digit binary string, which is a list of 1’s and 0’s. How many such byte strings are there? How many end in 0? How many have the second and fourth digit as 1’s?
2.
(a) Compute 3!, 4!, 5!, 10!.
(b) How many 9-digit numbers (using 1, 2, 3, ..., 9) are there if repetition is not allowed? How many are there which have at least one digit repeated?
(c) Compute \(\frac{100!}{98!}\).
The calculation shown is:
\[
= \frac{100 \cdot 99 \cdot 98!}{98!} = 100 \cdot 99 = 9900
\]
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