Solution to A. COUNTING METHODS (5 pts each) 1. How many strings of length 4 can be … - Sikademy
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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?

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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}


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.

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