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STD 11 - 6. permutation and combination — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 6. permutation and combination1 Mark
MCQ
$5 -$ digit numbers are to be formed using $2, 3, 5, 7, 9$ without repeating the digits. If $p$ be the number of such numbers that exceed $20000$ and $q$ be the number of those that lie between $30000$ and $90000$, then $p : q$ is
  • A
    $6:5$
  • B
    $3: 2$
  • C
    $4:3$
  • ✓
    $5: 3$
✓

Answer

Correct option: D.
$5: 3$
d
$p\,:\,\begin{array}{*{20}{c}}
  0&0&0&0&0 \\ 
  5&4&3&2&1 
\end{array}\,\,\,\begin{array}{*{20}{c}}
  {place} \\ 
  {ways} 
\end{array}$

Total no. of ways $=5!=120$

Since all numbers are $>20,000$

$\therefore $ all numbers $2,3,5,7,9$ can come at first place.

$q\,:\,\begin{array}{*{20}{c}}
  0&0&0&0&0 \\ 
  3&4&3&2&1 
\end{array}\,\,\,\begin{array}{*{20}{c}}
  {place} \\ 
  {ways} 
\end{array}$

Total no. of ways $=3 \times 4!=72$

( $\therefore $ $2$ and $9$ can not be put at first place)

So, $p :q=120 : 72=5 :3$

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