Answer to Question #191356 in Discrete Mathematics for Maricel

Question

A. COUNTING METHODS (5 pts each)

1. How many strings of length 4 can be formed using the letters ABCDE if it starts with letters AC and repetition is not

allowed?

2. There are 10 multiple choice questions in an examination. Each of the questions have four choices. In how many ways

can an examinee give possible answers?

B. BINOMIAL COEFFICIENTS

Expand (2𝑥 + 4𝑎) 4 using the binomial theorem. (10 pts)

C. PIGEONHOLE PRINCIPLE (5 pts) . Explain briefly.

Do you agree that there are 3 persons who have the same first and last name? Why and why not? 

Answer 1

(A)

1.Number of ways in which 4 length string formed when first two words are $AC= 1\times ^3C_2=3$

2.Number of ways in which examinee fives possible answer$= 4^{10}$

(B)

Given expression-

$(2x+4a)^4\\[9pt] =^4C_0(2x)^0(4a)^4+^4C_1(2x)^1(4a)^3+^4C_2(2x)^2(4a)^2+^4C_3(2x)^3(4a)^1+^4C_4(2x)^4(4a)^0 \\[9pt]=256a^4+512xa^3+384a^2x^2+32x^3a+46x^4$

(C) Yes, we can say that number of person who have the same first and last name is 3.

There are $(26\times 26)=676$ distinct combinations of first and last initials, with 676 students It is certainly possible due to assign each

students a distinct set of first and last initials. However by the peigenhole principle , having n=676, students we can say that There are 3 person with first and last name.