Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

There are $20$ boxes numbered $1,2,......20$. Five boxes are selected at random and are arranged in ascending order. In how many ways this can be done so that no. $10$ is present in each selection and it comes in $3rd$ place in the arrangement.?

$\begin{array}{1 1} (A) ^{10}C_2\times ^{10}C_2 \\ (B) ^9C_2\times ^{10}C_2 \\ (C) ^9C_2\times ^{9}C_2 \\ (D) ^{19}C_4 \end{array} $

Can you answer this question?

1 Answer

0 votes
Since no. 10 is present in each selection, only 4 boxes are to be selected.
Also since no. 10 comes in $3^{rd}$ position,
first two boxes should be numbered <10 and
next two numbers should be >10.
2 numbers are to be selected from 1,2.......9 in $^9C_2$ ways
2 numbers are to be selected from 11,12,......20 in $^{10}C_2$ ways.
$\therefore$ The required no. of arrangements = $^9C_2\times ^{10}C_2$


answered Sep 5, 2013 by rvidyagovindarajan_1
edited Dec 20, 2013 by meenakshi.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App