NEIGHBOUR DIGITS
Using the digits 0,1,2,3,4 find the number of 10 digit sequences that can be formed with the difference between any two consecutive digits equal to 1. CLARIFICATION:A sequence may start with 0.Look i said sequence not numbers.... Hope u will enjoy solving the problem :)
The answer is 648.
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1 solution
Indeed, we can show that P ( 2 n + 1 ) Q ( 2 n + 1 ) = = 2 × 3 n − 1 3 n P ( 2 n ) Q ( 2 n ) = = 3 n − 1 2 × 3 n − 1 and so the total number of sequence of length n is X ( n ) = 2 P ( n ) + 2 Q ( n ) + 2 Q ( n − 1 ) , and hence X ( 2 n + 1 ) = 1 4 × 3 n − 1 X ( 2 n ) = 8 × 3 n − 1 for any integer n ≥ 1 . Thus X ( 1 0 ) = 8 × 3 4 = 6 4 8 .
See if u replace x by (4-x) then u get a valid sequence. So let the no. of numbers of length n starting with 0 and 1 be P(n) & Q(n) respectively . So the no. of digits starting with 4 is P(n) and 3 is Q(n) too right??(1st Logic!!).Now a no. beginning with 2 has 1 or 3 as the next neighbour and so no. of such sequences(obviously of length n) is 2Q(n-1).Ok now a no. starting with 0 must have 1 and a no with 4 must have a no. 3 next....So we can claim P(n)=Q(n-1) right?? Ok now total number of such sequences is X(n)= P(n)+Q(n)+(2Q(n-1))+P(n)+Q(n)...(Each no. begins with 0/1/2/3/4...hence their sum).Ok using previous results X(n)=2P(n)+2Q(n-1)+2Q(n)=2Q(n)+4Q(n-1).Ok Look!! Q(n)=n length no. starting with 1=(n-1) lenth no. starting with 0 or 2=2Q(n-2)+P(n)=3Q(n-1)...Ok now it is mostly done....X(n)=2Q(n)+4Q(n-1)=2.3Q(n-2)+4.3Q(n-3)=......Go on reducing...Q(2)=2(10&12)&Q(1)=1....Finally u arrive Q(10)=2X3X3X3X3XQ(2)+4X3X3X3X3XQ(1)=648. ANS :)