(a) Explanation Proof: Let us we know that D n is given by, D n = n ! ( 1 − 1 1 ! + 1 2 ! − 1 3 ! + ... + ( − 1 ) n 1 n ! ) Therefore, we have D n − 1 = ( n − 1 ) ! ( 1 − 1 1 ! + 1 2 ! − 1 3 ! + ... + ( − 1 ) n − 1 1 ( n − 1 ) ! ) And = ( n − 2 ) ! ( 1 − 1 1 ! + 1 2 ! − 1 3 ! + ... + ( − 1 ) n − 2 1 ( n − 1 ) ! ) Now, ( n − 1 ) ( D n − 1 + D n − 2 ) = ( n − 1 ) { ( n − 1 ) ! ( 1 − 1 1 ! + 1 2 ! − 1 3 ! + ... + ( − 1 ) n − 1 1 ( n − 1 ) ! ) + ( n − 2 ) ! ( 1 − 1 1 ! + 1 2 ! − 1 3 ! + ... + ( − 1 ) n − 2 1 ( n − 2 ) ! ) } Continuation of the above, = [ n ( n − 1 ) ! ( 1 − 1 1 ! + 1 2 ! − 1 3 ! + (b) To determine To prove: The D n = n D n − 1 + ( − 1 ) n for all n ≥ 2. (c) To determine The D n a n d D n n ! to six decimal places for n = 1 , 2 , 3 , ... , 10.

Discrete Mathematics with Graph Theory (Classic Version) (3rd Edition) (Pearson Modern Classics for Advanced Mathematics Series)

3rd Edition

ISBN: 9780134689555

Author: Edgar Goodaire, Michael Parmenter

Publisher: PEARSON

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Chapter 7.6, Problem 11E

(a)

To determine

To prove: The Dn=(n−1)(Dn−1+Dn−2) for all n≥3.

(b)

To determine

To prove: The Dn=nDn−1+(−1)n for all n≥2.

(c)

To determine

The Dn and Dnn! to six decimal places for n=1,2,3,...,10.

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Chapter 7.6, Problem 10E

Chapter 7.6, Problem 12E

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Chapter 7 Solutions

Discrete Mathematics with Graph Theory (Classic Version) (3rd Edition) (Pearson Modern Classics for Advanced Mathematics Series)

Ch. 7.1 - True/False Questions 5!=120 Ch. 7.1 - Prob. 2TFQ Ch. 7.1 - Prob. 3TFQ Ch. 7.1 - Prob. 4TFQ Ch. 7.1 - Prob. 5TFQ Ch. 7.1 - Prob. 6TFQ Ch. 7.1 - True/False Questions 7. An r-permutation of n...Ch. 7.1 - Prob. 8TFQ Ch. 7.1 - Prob. 9TFQ Ch. 7.1 - True/False Questions There are 4!=24 ways in which...

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To prove: The D n = ( n − 1 ) ( D n − 1 + D n − 2 ) for all n ≥ 3.

Solution Summary: The author explains that the D_n=(n-1) and =-13!

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To prove: The Dn=(n−1)(Dn−1+Dn−2) for all n≥3.

To prove: The Dn=nDn−1+(−1)n for all n≥2.

The Dn and Dnn! to six decimal places for n=1,2,3,...,10.

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Chapter 7 Solutions