163 Discrete Mathematics Review 2 Use the following to answer ...

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$Math 171 Review for the Final Exam$

$Math 017 Materials With Exercises$

$Math 171 Test 2$

$MATH 171 SYLLABUS$

13. A market sells ten kinds of soda. You want to buy 12 bottles. How many possibilities are there if you want at most three bottles of any kind Solution. We will use the formula on page 506 and follow the pattern of Example 1 on the same page. Our problem can be reformulated as follows. How many solutions does the equation x 1+ … + x10 = 12 have, where x , , 1 … x1 0 are nonnegative integers and xi ≤ 3, i = 1, …,10 . Let Pi , i = 1, …,10, be the property that x i > 3. Notice that for any four distinct indices 1≤ i1 < i2 < i3 < i4 ≤10 we have N( Pi P ) 0 1 i 2 Pi 3 P i 4 = . Indeed, otherwise the number of bottles would be at least 16. Therefore in our case the formula on page 506 takes the form N( P … P ) = N − N( P) + N( PP) − N( PPP ). ' ' 1 10 ∑ ∑ ∑ i i j i 1≤≤ i 10 1≤the right part of the formula above. We can find the total number N of nonnegative integer solutions of the equation x1+ … + x10 = 12 by following the pattern in Example 5 on page 373. N = C(10 + 12 − 1,12) = C(21,12) = C(21,9) . Next, again by following Example 5 on page 373, we see that for any index i,1≤ i≤ 10, whence N P C C ' ( i ) = (10 + 8 − 1,8) = (17,8) Similarly for any and therefore, ∑ N( Pi ) = 10 ⋅C(17,8) . 1≤≤ i 10 i, j,1≤ i< j ≤10we have N( PP) = C(10 + 4 − 1, 4) = C(13, 4) i ∑ j 1≤answer is C(21,9) −10 ⋅ C(17,8) + C(10,2) ⋅C(13,4) − C(10,3) = 82885 . 14. How many permutations of all 26 letters of the alphabet are there, that contain at least one of the words SWORD, PLANT, CARTS Solution. Let A, B, and C be the sets of all permutations which contain SWORD, PLANT, or CARTS, respectively. Notice that A∩ C = B∩ C =∅. Therefore by inclusion – exclusion principle for three sets we have N( A∪B∪ C) = N( A) + N( B) + N( C) −N( A∩B). Let us compute N( A) . SWORD can start from position 1 up to position 22; in each case the remaining 21 letters can be arranged in 21! ways. Therefore N( A ) = 22⋅ 21! = 22! . Similarly NB ( ) = N( C= ) 22! Next we compute N( A∩ B) . Let us first look at the case when SWORD goes before PLANT. SWORD can start from any position from 1 to 17, the respective numbers of ways to place PLANT will be 17, 16,…,1. The remaining 16 letters can be arranged in 16! ways. There is the same number of ways to place PLANT in front of SWORD. Thus we obtain 17⋅18 N( A∩ B) = 2(1 + 2 + … + 17)16! = 2 16! = 18! 2 The answer is N( A∪B∪ C) = 3⋅22! − 18! = 3371995780959117312000. 15. Find the number of ways to climb a 12-step staircase, if you go up either one or three steps at a time. Solution. Let a n be the number of ways to climb n steps by going one or three steps at a time. Then we have a recurrent relation a = a + a n n−1 n−3 with the initial conditions a1 = a2 = 1, a3 = 2 . From here it is easy to compute directly that a 12 = 60 . Page 7
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