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Let-a-n-denote-the-number-of-all-n-digit-positive-integers-formed-by-the-digits-0-1-or-both-such-that-no-consecutive-digits-in-them-are-0-Let-b-n-the-number-of-such-n-digit-integers-ending-with-




Question Number 21913 by Tinkutara last updated on 06/Oct/17
Let a_n  denote the number of all n-digit  positive integers formed by the digits  0, 1 or both such that no consecutive  digits in them are 0. Let b_n  = the  number of such n-digit integers ending  with digit 1 and c_n  = the number of  such n-digit integers ending with  digit 0.  1. Which of the following is correct?  (1) a_(17)  = a_(16)  + a_(15)   (2) c_(17)  ≠ c_(16)  + c_(15)   (3) b_(17)  ≠ b_(16)  + c_(16)   (4) a_(17)  = c_(17)  + b_(16)   2. The value of b_6  is
$$\mathrm{Let}\:{a}_{{n}} \:\mathrm{denote}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{all}\:{n}-\mathrm{digit} \\ $$$$\mathrm{positive}\:\mathrm{integers}\:\mathrm{formed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{digits} \\ $$$$\mathrm{0},\:\mathrm{1}\:\mathrm{or}\:\mathrm{both}\:\mathrm{such}\:\mathrm{that}\:\mathrm{no}\:\mathrm{consecutive} \\ $$$$\mathrm{digits}\:\mathrm{in}\:\mathrm{them}\:\mathrm{are}\:\mathrm{0}.\:\mathrm{Let}\:{b}_{{n}} \:=\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{such}\:{n}-\mathrm{digit}\:\mathrm{integers}\:\mathrm{ending} \\ $$$$\mathrm{with}\:\mathrm{digit}\:\mathrm{1}\:\mathrm{and}\:{c}_{{n}} \:=\:\mathrm{the}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{such}\:{n}-\mathrm{digit}\:\mathrm{integers}\:\mathrm{ending}\:\mathrm{with} \\ $$$$\mathrm{digit}\:\mathrm{0}. \\ $$$$\mathrm{1}.\:\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{correct}? \\ $$$$\left(\mathrm{1}\right)\:{a}_{\mathrm{17}} \:=\:{a}_{\mathrm{16}} \:+\:{a}_{\mathrm{15}} \\ $$$$\left(\mathrm{2}\right)\:{c}_{\mathrm{17}} \:\neq\:{c}_{\mathrm{16}} \:+\:{c}_{\mathrm{15}} \\ $$$$\left(\mathrm{3}\right)\:{b}_{\mathrm{17}} \:\neq\:{b}_{\mathrm{16}} \:+\:{c}_{\mathrm{16}} \\ $$$$\left(\mathrm{4}\right)\:{a}_{\mathrm{17}} \:=\:{c}_{\mathrm{17}} \:+\:{b}_{\mathrm{16}} \\ $$$$\mathrm{2}.\:\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{b}_{\mathrm{6}} \:\mathrm{is} \\ $$

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