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1.28. Evaluate (a) 7 C4, (b) 6 C5, (c) 4 C4.
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1.29. In how many ways can a committee of 5 people be chosen out of 9 people 9 5
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1.30. Out of 5 mathematicians and 7 physicists, a committee consisting of 2 mathematicians and 3 physicists is to be formed. In how many ways can this be done if (a) any mathematician and any physicist can be included, (b) one particular physicist must be on the committee, (c) two particular mathematicians cannot be on the committee
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(a) 2 mathematicians out of 5 can be selected in 5C2 ways. 3 physicists out of 7 can be selected in 7C3 ways. Total number of possible selections (b) 2 mathematicians out of 5 can be selected in 5C2 ways. 2 physicists out of 6 can be selected in 6C2 ways. Total number of possible selections (c) 2 mathematicians out of 3 can be selected in 3C2 ways. 3 physicists out of 7 can be selected in 7C3 ways. Total number of possible selections
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10 35
6C2
10 15
7C3
3 35
CHAPTER 1 Basic Probability
1.31. How many different salads can be made from lettuce, escarole, endive, watercress, and chicory
Each green can be dealt with in 2 ways, as it can be chosen or not chosen. Since each of the 2 ways of dealing with a green is associated with 2 ways of dealing with each of the other greens, the number of ways of dealing with the 5 greens 25 ways. But 25 ways includes the case in which no greens is chosen. Hence, Number of salads 25 1 31
Another method
One can select either 1 out of 5 greens, 2 out of 5 greens, . . . , 5 out of 5 greens. Then the required number of salads is
5C1 5C2 5C3 5C4 nC2 5C5 nC3
10 c
5 2n
1 1.
In general, for any positive integer n, nC1
1.32. From 7 consonants and 5 vowels, how many words can be formed consisting of 4 different consonants and 3 different vowels The words need not have meaning.
The 4 different consonants can be selected in 7C4 ways, the 3 different vowels can be selected in 5C3 ways, and the resulting 7 different letters (4 consonants, 3 vowels) can then be arranged among themselves in 7 P7 7!
ways. Then
Number of words
5C3 7!
35 10 5040
1,764,000
The Binomial Coefficients
1.33. Prove that
n r We have n r n r 1
1 . 1 r)!
r!(n
n(n r!(n (n
1)! r)!
r r)(n 1)! r!(n r)! r(n r!(n (r 1)! r)!
r)(n 1)! r!(n r)!
(n 1)! r!(n r 1)! n r 1 n r
The result has the following interesting application. If we write out the coefficients in the binomial expansion of (x y)n for n 0, 1, 2, . . . , we obtain the following arrangement, called Pascal s triangle: n 0 n 1 n 2 n 3 n 4 n 5 n 6 etc. 1 1 1 1 1 1 1 6 5 15 4 10 20 3 6 10 15 2 3 4 5 6 1 1 1 1 1 1
1 1
(n 1)! 1)!(n r)!
An entry in any line can be obtained by adding the two entries in the preceding line that are to its immediate left and right. Therefore, 10 4 6, 15 10 5, etc.
x 2 1 x .
CHAPTER 1 Basic Probability
1.34. Find the constant term in the expansion of
According to the binomial theorem, x2 1 x
The constant term corresponds to the one for which 3k 12 4
12 1 (x 2)k x k 12
12 k
0, i.e., k
12 11 10 9 4 3 2 1
12 12 a k x 3k k 0
4, and is therefore given by
Probability using combinational analysis 1.35. A box contains 8 red, 3 white, and 9 blue balls. If 3 balls are drawn at random without replacement, determine the probability that (a) all 3 are red, (b) all 3 are white, (c) 2 are red and 1 is white, (d) at least 1 is white, (e) 1 of each color is drawn, (f) the balls are drawn in the order red, white, blue.
(a) Method 1
Let R1, R2, R3 denote the events, red ball on 1st draw, red ball on 2nd draw, red ball on 3rd draw, respectively. Then R1 > R2 > R3 denotes the event all 3 balls drawn are red. We therefore have P(R1 > R2 > R3) P(R1) P(R2 u R1) P(R3 u R1 > R2) 8 7 6 20 19 18 14 285
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