1. **Problem statement:** We want to find the number of distinct paths from point $(0,0)$ to point $(3,3)$ on a grid, where movement is allowed only to the right, upward, or diagonally upward-right. 2. **Understanding the moves:** - Right move: $(x,y) \to (x+1,y)$ - Upward move: $(x,y) \to (x,y+1)$ - Diagonal move: $(x,y) \to (x+1,y+1)$ 3. **Key insight:** Each path consists of a sequence of moves that increase the $x$ coordinate by 3 in total and the $y$ coordinate by 3 in total. 4. **Let:** - $r$ = number of right moves - $u$ = number of upward moves - $d$ = number of diagonal moves 5. **From the moves, we have the system:** $$r + d = 3$$ $$u + d = 3$$ 6. **Solving for $r$ and $u$ in terms of $d$:** $$r = 3 - d$$ $$u = 3 - d$$ 7. **Since $r,u,d \geq 0$, $d$ can be $0,1,2,$ or $3$. 8. For each fixed $d$, the total number of moves is:** $$n = r + u + d = (3 - d) + (3 - d) + d = 6 - d$$ 9. **Number of distinct sequences for fixed $d$ is the multinomial coefficient:** $$\frac{n!}{r!\,u!\,d!} = \frac{(6 - d)!}{(3 - d)! (3 - d)! d!}$$ 10. **Calculate for each $d$ and sum:** - For $d=0$: $\frac{6!}{3!3!0!} = \frac{720}{6 \times 6} = 20$ - For $d=1$: $\frac{5!}{2!2!1!} = \frac{120}{2 \times 2 \times 1} = 30$ - For $d=2$: $\frac{4!}{1!1!2!} = \frac{24}{1 \times 1 \times 2} = 12$ - For $d=3$: $\frac{3!}{0!0!3!} = \frac{6}{1 \times 1 \times 6} = 1$ 11. **Total number of distinct paths:** $$20 + 30 + 12 + 1 = 63$$ **Final answer:** There are **63** distinct paths from $(0,0)$ to $(3,3)$ with the allowed moves.