Question

How many 7 digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2, 4?

• A. 120
• B. 360
• C. 240
• D. 424

Answer 1

The correct option is B 360

There are 7 digits 1,2,0,2,4,2,4 in which 2 occurs 3 times, 4 occurs 2 times.
$\therefore$ Number of 7 digits numbers $=\frac{7!}{3!\times 2!}=420$
But out of these 420 numbers, there are some numbers which being with '0' and they are not 7 digit numbers. The number of such numbers beginning with '0'.
$=\frac{6!}{3!\times 2!}=60$
Hence the required number of 7 digits numbers
=420 - 60 = 360
Alternatively:
$\begin{array}{ccccccc}6& 6& 5& 4& 3& 2& 1\end{array}\Rightarrow 6.(6!)$
Since the leftmost place can assume only non-zero digit hence it can consider only 6 digits. Noe the remaining 6 places can be filled up in normal ways.
But since 2 occurs 3 times, 4 occurs 2 times, therefore required number of 7 digit numbers $=\frac{6.(6!)}{3!.2!}=360$