Question

How many numbers greater than 10 lakhs be formed from 2, 3, 0, 3, 4, 2, 3.
$\left( A \right)$ 420
$\left( B \right)$ 360
$\left( C \right)$ 400
$\left( D \right)$ 300

Answer 1

Hint – In this question use the concept that 10 lakhs = 1,000,000 consisting of 7 digits and the given digits are also 7 so the number cannot start from zero otherwise it will be less than 10 lakhs or it will convert into 6 digit number so fixed the first digit and arrange the remaining digits so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given number:
10 Lakhs
As we all know 10 lakhs = 1,000,000 so it consists of 7 digits.
Now the given digits are 2, 3, 0, 3, 4, 2, and 3.
So we have to form the numbers greater than 10 lakhs.
So as we see that in the given digits there is no one and the number of digits are 7 so the number starts from any given digits except zero, otherwise it will convert into a 6 digit number which is less than 10 lakhs.
So there are following cases:
$\left( i \right)$ When the first digit is two.
So the number of ways to arrange the remaining 6 digits are $\dfrac{{6!}}{{3!}}$ as there are three 3's present in the remaining digits so we have to divide by 3!
$\left( {ii} \right)$ When the first digit is three.
So the number of ways to arrange the remaining 6 digits are $\dfrac{{6!}}{{2!.2!}}$ as there are two 3’s and two 2’s are present in the remaining digits so we have to divide by (2!).(2!)
$\left( {iii} \right)$ When the first digit is four.
So the number of ways to arrange the remaining 6 digits are $\dfrac{{6!}}{{3!.2!}}$ as there are three 3’s and two 2’s are present in the remaining digits so we have to divide by (3!).(2!)
So the total numbers greater than 10 lakhs are the sum of all the above cases.
So the total numbers greater than 10 lakhs are = $\dfrac{{6!}}{{3!}} + \dfrac{{6!}}{{2!.2!}} + \dfrac{{6!}}{{3!.2!}}$
Now as we know that 6! = 720 and 3! = 6 and 2! = 2 so use these values in the above equation we have,
So the total numbers greater than 10 lakhs are = $\dfrac{{720}}{6} + \dfrac{{720}}{{2\left( 2 \right)}} + \dfrac{{720}}{{6\left( 2 \right)}}$
Now simplify this we have,
$\Rightarrow 120 + 180 + 60 = 360$
So this is the required answer.
Hence option (B) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember if there are n digits in the system in which r digits are same then the number of ways to arrange them is $\dfrac{{n!}}{{r!}}$, if there are two types of digits repeated (i.e. one type of r digits are same and another type of p digits are same), then the number of ways to arrange them is $\dfrac{{n!}}{{r!.p!}}$.