How to Select Number

We change the prob to the correspondence prob such that, Let $B=\{1,2,3,...,10\}$ we are going to select three number such that no two consecutive number. We will set the selected number $a_1<a_2<a_3$ ; $1<a_i<10$ but in this case there would be $(1,2,3), etc.$ among the selected number. So we take $b_1=a_1$ ; $b_2=a_2+1$ ; $b_3=a_3+2$ so that there are no two consecutive number. So now we can choose freely $b_1,b_2,b_3$ from $1<b_i<8$ The answer is $C^{8}_3=56$

$\begin{aligned}A&=\{1, 3, 5, \dots,19\}\hspace{4mm}\color{#3D99F6}\small \text{(Given)}\\\\ &\text{Divide A into } 2\text{ sets B and C such that,}\\\\ B&=\{1, 5, 9,13,17\}\\ C&=\{3, 7, 11,15,19\}\\ &\text{B and C are complementary sets under A}\\\\ &\text{The only way to pick } a_{1},a_{2},a_{3} \text{ so that } \lvert{a_{i}-a_{j}\rvert\neq2}\text{ is,}\\\\ &\underline{\text{Case-1:}}\hspace{2mm}\text{When all 3 elements are picked from one of the sets B or C}\\\\ &\text{This can be done in}\\ &^5C_3+^5C_3 =20 \text{ ways}\\\\ &\underline{\text{Case-2:}}\hspace{2mm}\text{When 1 element is picked from one of the sets and 2 from the other one.}\\\\ &\text{Since,}\lvert{a_{i}-a_{j}\rvert\neq2}\\ \implies&\text{If an element is picked you cannot pick any of its neighbors in the original set A}\\ &\text{Also,an element and its neighbor cannot fall in the same set B or C}\\\\ &1,19 \text{ has only 1 neighbor in the complementary set.}\\ &\text{If 1 or 19 was chosen as the first element ,then}\\ &\text{There are } ^4C_2 \text{ways to choose the other 2 elements from the complementary set.}\\ &\text{Thus triples involving 1 or 19 as the first element can be chosen in } 2\times^4C_2=12 \text{ ways}\\\\ &\text{All elements other than1,19} \text{ has 2 neighbors in the complementary set.}\\ &\text{If one of the elements other than 1,19 was chosen as the first element ,then}\\ &\text{There are } ^3C_2 \text{ways to choose the remaining 2 elements from the complementary set.}\\ &\text{Thus triples not involving 1,19 as the first element can be chosen in } 8\times^3C_2=24 \text{ ways}\\\\ &\text{Thus the total number of ways for choosing 3 elements is given by,}\\ &\text{Number of ways}=20+12+24\\ &\hspace{31mm}=\color{#EC7300}\boxed{\color{#333333}56} \color{#333333}\text{ ways}\end{aligned}$

1 8 28 56 1 7 21 35 1 6 15 20 1 5 10 10 1 4 6 4 1 3 3 1 1 2 1 1 1