1in A Class Of50students Everyone Has Either A Pierced Nose Or A P ✓ Solved

1. In a class of 50 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose to raise his or her hand. Nine hands go up. Then the professor asked everyone with a pierced ear to do likewise.

This time there are 44 hands raised. How many students have piercings both on their ears and their noses? 2. Give a recurrence relation that describes the sequence 2, 4, 8, 16, 32, 64, 128, .... P ( n ) = if n = 1 · P ( n − 1) if n > .

The following problem refers to strings in A, B, ..., Z. How many four-letter strings are there that begin with S ? 4. Form a seven-letter word by mixing up the letters in the word COMBINE. (a) How many ways can you do this? (b) How many ways can you do this if all the vowels have to be at the beginning? (c) How many ways can you do this if no vowel is isolated between two consonants? 5. How many solutions (using only nonnegative integers) are there to the following equation? x 1 + x 2 + x 3 + x 4 + x 5 = 36

Paper for above instructions

Solutions to the Problems

Problem 1: Pierced Noses and Ears

In a class of 50 students, each student has either a pierced nose or a pierced ear. We know that:
- 9 students have a pierced nose (let’s denote this as \(N\)).
- 44 students have a pierced ear (let’s denote this as \(E\)).
- The total number of students is 50.
We aim to determine how many students have both piercings (denote this as \(X\)), using the principle of inclusion-exclusion. The formula for our scenario can be expressed as follows:
\[
N + E - X = \text{Total Students}
\]
Substituting the known values into the equation:
\[
9 + 44 - X = 50
\]
This simplifies to:
\[
53 - X = 50
\]
Rearranging gives us:
\[
X = 3
\]
Thus, 3 students have piercings on both their ears and noses.

Problem 2: Recurrence Relation for the Sequence

The sequence \(2, 4, 8, 16, 32, 64, 128, \ldots\) is a geometric sequence where each term is 2 times the previous term. We can express the \(n^{th}\) term, \(P(n)\), of this sequence with the following relation:
Base case:
- \(P(1) = 2\)
Recurrence relation:
- \(P(n) = 2 \cdot P(n - 1)\) for \(n > 1\)
This means that every term can be obtained by multiplying the previous term by 2.

Problem 3: Four-Letter Strings Beginning with 'S'

To find how many four-letter strings can be formed starting with 'S', we consider the following:
1. The first letter is already determined as 'S'.
2. The remaining three letters can be any of the 26 letters in the alphabet.
As the remaining 3 letters can each be any of the 26 alphabets, the total number of strings can be calculated by:
\[
26 \times 26 \times 26 = 26^3 = 17576
\]
Thus, there are 17,576 four-letter strings that begin with 'S'.

Problem 4: Arrangements of the Word "COMBINE"

(a) To determine how many ways we can arrange the letters in "COMBINE", we first note that "COMBINE" consists of 7 distinct letters.
\[
\text{Number of arrangements} = 7! = 5040
\]
Therefore, there are 5,040 ways to arrange the letters in "COMBINE".
(b) If we require all the vowels (O, I, E) to be at the beginning, we can arrange the 3 vowels in the first three positions and the remaining consonants (C, M, B, N) in the last four positions.
The arrangements for the vowels are:
\[
3! = 6
\]
The arrangements for the consonants are:
\[
4! = 24
\]
Thus, the total arrangements where vowels are at the beginning is:
\[
3! \times 4! = 6 \times 24 = 144
\]
Hence, there are 144 arrangements of "COMBINE" where all vowels come first.
(c) For the requirement that no vowel is isolated between two consonants, we can use grouping.
We can consider vowels as a block. Thus, for the arrangement of consonants and the block of vowels, we can have:
1. Arrangement of the consonants C, M, B, N can be arranged with the vowel block (OIE).
Assuming the vowels form a block \(X\), we can arrange \(C, M, B, N, X\):
\[
5! = 120 \text{ (arranging 5 units)}
\]
Each arrangement of vowels within the block can occur in:
\[
3! = 6 \text{ (arranging O, I, E)}
\]
Thus, the total number of arrangements satisfying this condition is:
\[
5! \times 3! = 120 \times 6 = 720
\]
So, there are 720 arrangements of "COMBINE" where no vowel is isolated between consonants.

Problem 5: Nonnegative Integer Solutions

To find how many nonnegative integer solutions exist for the equation
\[
x_1 + x_2 + x_3 + x_4 + x_5 = 36,
\]
we can apply the stars and bars theorem, where the number of solutions to the equation \(x_1 + x_2 + ... + x_k = n\) in nonnegative integers is given by:
\[
\binom{n + k - 1}{k - 1}.
\]
In this equation, \(n = 36\) and \(k = 5\):
\[
\text{Number of solutions} = \binom{36 + 5 - 1}{5 - 1} = \binom{40}{4}.
\]
Calculating this:
\[
\binom{40}{4} = \frac{40 \times 39 \times 38 \times 37}{4 \times 3 \times 2 \times 1} = 91,390.
\]
Thus, there are 91,390 nonnegative integer solutions to the equation.

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