First slide

Permutations

Question

The number of numbers greater than 10⁶ that can be formed using the digits of the number 2334203, if all the digits of the given number must be used, is

Moderate

A

360
B

420
C

260
D

None of these

Solution

The total number of permutations of the digits of given number is
$\frac{7!}{2! 3!} = 420$
But when 0 occupies the first position, the number becomes a six-digit number which are not acceptable, since such a number will become less then 10⁶.
The total number of such six-digit numbers is
$= \frac{6!}{2! 3!} = 60$
Hence, the required number is 420 – 60 = 360.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App