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Standard VII

Mathematics

Fundamental Principle of Counting

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Question

Find the total possible integral solution for $n_{1} n_{2} = 2 n_{1} - n_{2}$ ,where $n_{1}, n_{2} ϵ$ integer.

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Solution

$n_{1} n_{2} = 2 n_{1} - n_{2}$
$\Rightarrow n_{2} (n_{1} + 1) = 2 n_{1} \Rightarrow n_{2} = \frac{2 n_{1}}{n_{1} + 1}$
or $n_{2} = 2 - \frac{2}{n_{1} + 1}$ ; Since, $n_{1}, n_{2}$ is an integer.
$∴ \frac{2}{n_{1} + 1} \in i n t e g e r$
or $n_{1} + 1 = - 2, - 1, 1, 2$
or $n_{1} = - 3, - 2, 0, 1 \Rightarrow n_{2} = 3, 4, 0, 1$
$\Rightarrow$ Integral solutions of the form $(n_{1}, n_{2})$ are $(- 3, 3), (- 2, 4), (0, 0), (1, 1)$ .

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Find the total possible integral solution for n1n2=2n1−n2,where n1,n2ϵ integer.

n1n2=2n1−n2⇒n2(n1+1)=2n1⇒n2=2n1n1+1or n2=2−2n1+1; Since, n1,n2 is an integer.∴2n1+1∈integeror n1+1=−2,−1,1,2or n1=−3,−2,0,1⇒n2=3,4,0,1⇒ Integral solutions of the form (n1,n2) are (−3,3),(−2,4),(0,0),(1,1).

Find the total possible integral solution for $n_{1} n_{2} = 2 n_{1} - n_{2}$ ,where $n_{1}, n_{2} ϵ$ integer.