2010 AMC 8 Problem 25 Recursion – Combinatorics

Understand the problem

Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$, $2$, or $3$ at a time. For example, Jo could climb $3$, then $1$, then $2$. In how many ways can Jo climb the stairs? $\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$
Source of the problem

2010 AMC 8 Problem 25

Topic
Recursion – Combinatorics
Difficulty Level
Easy
Difficulty Level
Easy
Suggested Book
Excursion in Mathematics

Start with hints

Do you really need a hint? Try it first!

Try to recall the concept of recursion . Then proceed to apply in this question .  https://en.wikipedia.org/wiki/Recursion#In_mathematics  
The number of ways to climb one stair is $1$. \( \\ \) There are $2$ ways to climb two stairs: $1$,$1$ or $2$. \( \\ \) For 3 stairs, there are $4$ ways: ($1$,$1$,$1$) ($1$,$2$) ($2$,$1$) ($3$) . \( \\ \) Try to aply this for 4 stairs in recursive manner .
For  4 stairs : After climbing 1 stair in one step  , Jo can climb the rest 3 stairs in (1 , 1 , 1) ,(1 , 2), (2, 1) ,(3) i.e. in 4 ways . \( \\ \) After climbing 2 stairs in one step , Jo can climb the rest 2 stairs in (1 , 1 ) and ( 2)   i.e. in 2 ways . \( \\ \)   After climbing 3 stairs in one step , Jo can climb the rest 1 stair in  a  single step. \( \\ \) So to climb 4 stairs Jo can have (1+ 2 + 4) = 7 ways . Now time arrives to apply the recursion  so proceed to apply .        
Why will we apply recursion ? \( \\ \) Notice that to climb 5 stairs , \( \\ \) After climbing 1 stair in one step , we again have to find how many ways Jo can follow to climb the rest 4 stairs . \( \\ \) After climbing 2 stairs in one step , we again have to find how many ways Jo can follow to climb the rest 3 stairs . \( \\ \) After climbing 3 stairs in one step ,we again have to find how many ways Jo can follow to climb the rest 2 stairs . \( \\ \) So the no ways Jo can have for 5 stairs is  ( 7 + 4 + 2) = 13 . \( \\ \) So for 6 stairs , the no ways Jo has (13 + 7 +4 ) = 24 .

Watch the video

Connected Program at Cheenta

Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

Similar Problems

Colour Problem | PRMO-2018 | Problem No-27

Try this beautiful Problem on Combinatorics from PRMO -2018.You may use sequential hints to solve the problem.

Right-angled shaped field | AMC 10A, 2018 | Problem No 23

Try this beautiful Problem on triangle from AMC 10A, 2018. Problem-23. You may use sequential hints to solve the problem.

Area of region | AMC 10B, 2016| Problem No 21

Try this beautiful Problem on Geometry on Circle from AMC 10B, 2016. Problem-20. You may use sequential hints to solve the problem.

Coin Toss Problem | AMC 10A, 2017| Problem No 18

Try this beautiful Problem on Probability from AMC 10A, 2017. Problem-18, You may use sequential hints to solve the problem.

GCF & Rectangle | AMC 10A, 2016| Problem No 19

Try this beautiful Problem on Geometry on Rectangle from AMC 10A, 2010. Problem-19. You may use sequential hints to solve the problem.

Fly trapped inside cubical box | AMC 10A, 2010| Problem No 20

Try this beautiful Problem on Geometry on cube from AMC 10A, 2010. Problem-20. You may use sequential hints to solve the problem.

Measure of angle | AMC 10A, 2019| Problem No 13

Try this beautiful Problem on Geometry from AMC 10A, 2019.Problem-13. You may use sequential hints to solve the problem.

Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Geometry from PRMO -2018.You may use sequential hints to solve the problem.

Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.