Question

In a game, a person $P$ at point $(x,y)$ can jump to any one of the following points respectively: $(x+1,y),(x+2,y),(x,y+1),(x,y+2).$ If $P$ is at $(0,0)$, then he can reach:

• A. $(2,2)$ in $14$ ways
• B. $(2,3)$ in $32$ ways
• C. $(2,2)$ in $9$ ways
• D. $(2,3)$ in $26$ ways

Answer 1

Answer: (A), (B)

The correct options are
A $(2,2)$ in $14$ ways
B $(2,3)$ in $32$ ways

Let's denote the respective jumps as $L_1,L_2,U_1,U_2.$

For $(2,2):$
$L_1{L}_1{U}_1{U}_1\to \frac{4!}{2!2!}{L}_1{L}_1{U}_2\to \frac{3!}{2!}{L}_2{U}_1{U}_1\to \frac{3!}{2!}{L}_2{U}_2\to 2!$

Total $=6+3+3+2=14$ ways

For $(2,3):$
$L_1{L}_1{U}_1{U}_1{U}_1\to \frac{5!}{3!2!}{L}_1{L}_1{U}_1{U}_2\to \frac{4!}{2!}{L}_2{U}_1{U}_1{U}_1\to \frac{4!}{3!}{L}_2{U}_1{U}_2\to 3!$

Total $=10+12+4+6=32$ ways