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Permutations and Combinations (p-2) — Maths (commerce) STD 11 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Permutations and Combinations (p-2)3 Marks
Question
Find $m$ and $n$ if ${ }^{(m+n)} P_2=56$ and ${ }^{(m-n)} P_2=12$
✓

Answer

$
\begin{aligned}
& \mathrm{m}^n \mathrm{P}_2=56 \\
& \therefore \frac{(\mathrm{m}+\mathrm{n}) !}{(\mathrm{m}+\mathrm{n}-2) !}=56 \\
& \therefore \frac{(m+n)(m+n-1)(m+n-2) !}{(m+n-2) !}=56 \\
& \therefore(\mathrm{m}+\mathrm{n})(\mathrm{m}+\mathrm{n}-1)=8 \times 7
\end{aligned}
$
Comparing on both sides, we get
$
\mathrm{m}+\mathrm{n}=8 \text {.....(i) }
$
Also ${ }^{\mathrm{m}-\mathrm{n}} \mathrm{P}_2=12$
$
\begin{aligned}
& \therefore \frac{(m-n) !}{(m-n-2) !}=12 \\
& \therefore \frac{(m-n)(m-n-1)(m-n-2) !}{(m-n-2) !}=12 \\
& \therefore(m-n)(m-n-1)=4 \times 3
\end{aligned}
$
Comparing on both sides, we get
$
m-n=4
$
Adding (i) and (ii), we get
$
\begin{aligned}
& 2 \mathrm{~m}=12 \\
& \therefore \mathrm{m}=6
\end{aligned}
$
Substituting $m=6$ in (ii), we get
$
\begin{aligned}
& 6-n=4 \\
& \therefore n=2
\end{aligned}
$

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