Mymensingh Higher Secondry 2018#10

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samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm
Mymensingh Higher Secondry 2018#10

Unread post by samiul_samin » Mon Mar 12, 2018 8:27 pm

There are $20$ boxes on a table.There are $2$ balls in the $1$st box,$3$ balls in the $2$nd box,$4$ balls in the $3$rd box,.........,$21$ balls in the $20$th box.Bijoy wants to pick $2$ balls from any particular box. In how many ways he can do this work?

samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm

Re: Mymensingh Higher Secondry 2018#10

Unread post by samiul_samin » Mon Mar 12, 2018 8:40 pm

Answer
$\fbox {1540}$
Solution
We have to use $\dbinom nr$ formula and Additional Principle.
The total way of choosing is

$=\dbinom 22+\dbinom 32+\dbinom 42+\dbinom 52+\dbinom 62+\dbinom 72+\dbinom 82+\dbinom 92+\dbinom {10}2+\dbinom {11}2+\dbinom {12}2+\dbinom {13}2+\dbinom {14}2+\dbinom {15}2+\dbinom {16}2+\dbinom {17}2 + \dbinom {18}2+\dbinom {19}2+\dbinom {20}2+\dbinom {21}2=1540$

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