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# 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? $\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

# 2010 AMC 8 Problem 25

##### Topic
Recursion – Combinatorics
Easy
Easy
##### Suggested Book
Excursion in Mathematics

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 .

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