Consider a permutation (a₁, a₂, a₃, a₄, a₅ ) of 1,2,3,4,5 . We say the 5-tuple (a₁, a₂, a₃, a₄, a₅ ) is flawless if for all 1 i < j < k 5 , the sequence (a_j, a_j, a_k ) is not an arithmetic progression (in that order). Find the number of flawless 5-tuples.

Step-by-step solution

There are only four possible three terms AP exist, which are , , and Consider the sets A= are present in the sequence B= are present in the sequence C= are present in the sequence D= are present in the sequence aligned & n(A)=n(B)=n(C)=n(D)= ^5 C₃ 2 2 =40 & n(A B)=n(B C)= ^5 C₄ 2=10 & n(A C)=1+6 2=13 & n(A D)= ^5 C₄ 2=n(C D)=10 & n(B D)=4 |2| 2=16 & n(A B C)=2 & n(B C D)=4 & n(A C D)=2 & n(A B D)=4 and n(A B C D)=2 & Now n(A B C D)=4(40)-69+12-2=101 & So number of flawless 5-tuples =5-101=19 aligned