Forming NUMBERS...

Probability Level 3

How many numbers greater than 1,000,000 can be formed using each of the digits 2, 3, 0, 3, 4, 2, 3 exactly once?

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300 420 400 360

Harshvardhan Mehta by Harshvardhan Mehta , 23, India

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3 solutions

Vighnesh Raut
Jul 27, 2014

Total numbers that can be formed are 7!.

Numbers beginning with 0 are 6!

So, total numbers greater than 10 lakhs are

= 7 ! − 6 ! 2 ! 3 ! \frac { 7!-6! }{ 2!3! }

= 4320 12 \frac { 4320 }{ 12 }

=360

We divide by 2! and 3! because there are two 2's and three 3's.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

gud and an easy solution... good work

Harshvardhan Mehta - 6 years, 10 months ago

Since the umber must be greater than 10 lakhs, i.e., 10,00,000 it must be of seven digits and should begin with 2 or 3 or 4 .

No. of numbers beginning with 2. = 6 ! 3 ! = 120 = \frac{6!}{3!} = 120

No. of numbers beginning with 3. = 6 ! 2 ! . 2 ! = 180 = \frac{6!}{2!.2!} = 180

No. of numbers beginning with 4. = 6 ! 3 ! . 2 ! = 60 = \frac{6!}{3!.2!} = 60

Required number is 120 + 180 + 60 = 360

3 Helpful 0 Interesting 0 Brilliant 0 Confused

What is the explanation for 6 ! 3 ! \frac{6!}{3!}

Navho Sarker - 6 years, 5 months ago

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well 6! because total number of digits. and we have divided it by 3! because there are 3 2's.

Harshvardhan Mehta - 6 years, 4 months ago

Lets keep 0 aside for a moment.. So now number of possible arrangements of the digits is 6!/ (2! * 3!). So now the number of slots of fill in zero are 6 so total number of digits =6!/ (2! * 3!) *6=360

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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