Chapter 2 - Chapter 2 Question:2 How many 2n-digit positive...

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Chapter 2 Question:2 How many 2n -digit positive integers can be formed if the digits in odd positions (counting the rightmost digit as position 1 ) must be odd and the digits in even positions must be even and positive? Answer: In the (2n)-digit positive integer, there are n odd positions that must be filled with digits from the set 0, 3.5,7,91 and there are n even positions that must be filled with digits from the set {2, 4, 6, 8}, (since 0 isn't positive). Repetitions are allowed, and order matters. Thus, we have x = (20)" possibilities. Question:4 How many ternary strings of length 2n are there in which the zeroes appear only in odd-numbered positions? Answer: In the question, given: There is a ternary string of length 2N Possibilities: 0,1,2 Since the length is 2n, we have odd numbered position have length of 2 and even numbered position length n. Since zeros appear only in odd numbered position, it will have three posiisbilties 0,1,2 And so, even numbered positions will have only two possibilities 1 and 2. Number of ternary strings of length 2n which has zero only in odd positions. =(odd possibilities) length and (Even Possibilities) length = 3 n and 2 n =6 n Question : 6 Mrs. Ste ff en’s third grade class has 30 students in it. The students are divided into three groups (numbered 1, 2, and 3), each having 10 students. a) The students in group 1 earned 10 extra minutes of recess by winning a class competition. Before going out for their extra recess time, they form a single file line. In how many ways can they line up? b) When all 30 students come in from recess together, they again form a single file line. However, this time the students are arranged so that the first student is from group 1, the second from group 2, the third from group 3, and from there on, the students continue to alternate by group in this order. In how many ways can they line up to come in from recess? Answer: Given: Number of students =30 students. They were divided by 3 groups like GI, G2 and G3 Each group contains 10 Students. (a)
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They can be arranged in 10/30 ways. n! P (n, r) = n!/(n — r) P(10,10) =10! / (10 —10)! =10! =3628800 Therefore, they can be arranged in 13,628,800 ways. (b) * Given that when all 30 students come in from recess together, they again form a single file line-* The students are arranged so that the first student is from G. the second from G2, the third from G3, and from there on, the students continue to alternate by group in this order. G11,G21,G31, G12, G22, G13, G23,G33……..G110,G210,G310 Ist Student is from GI =10C1 =10 ways. Ist Student is from G2. =10 ways. Ist Student is from GI =10C1 =10 ways. Second Student is from GI = 9C1= 9 ways, Second Student is from G2 = 9C1 =9 ways.
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