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A path of length n is a sequence of poin...

A path of length `n` is a sequence of points `(x_(1),y_(1))`, `(x_(2),y_(2))`,….,`(x_(n),y_(n))` with integer coordinates such that for all `i` between `1` and `n-1` both inclusive,
either `x_(i+1)=x_(i)+1 ` and `y_(i+1)=y_(i)` (in which case we say the `i^(th)` step is rightward)
or `x_(i+1)=x_(i)` and `y_(i+1)=y_(i)+1` ( in which case we say that the `i^(th)` step is upward ).
This path is said to start at `(x_(1),y_(1))` and end at `(x_(n),y_(n))`. Let `P(a,b)`, for `a` and `b` non-negative integers, denotes the number of paths that start at `(0,0)` and end at `(a,b)`.
The sum `P(43,4)+sum_(j=1)^(5)P(49-j,3)` is equal to

A

`P(4,48)`

B

`P(3,49)`

C

`P(4,47)`

D

`P(5,47)`

Text Solution

Verified by Experts

The correct Answer is:
A

`(a)` `P(i,n-i)=^(n)C_(i)=P(n-i,i)`
Now `P(43,4)+sum_(j=1)^(5)P(49-j,3)`
`=('^(47)C_(4)+^(47)C_(3))+^(48)C_(3)+^(49)C_(3)+^(50)C_(3)+^(51)C_(3)`
`=('^(48)C_(4)+^(48)C_(3))+^(49)C_(3)+^(50)C_(3)+^(51)C_(3)`
`=('^(48)C_(4)+^(49)C_(3))+^(50)C_(3)+^(51)C_(3)=('^(50)C_(4)+^(50)C_(3))`
`=^(51)C_(4)+^(51)C_(3)=^(52)C_(4)=P(48,4)=P(4,48)`
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