How many 7-digit numbers can be arranged from 6690690

Question

How many 7-digit numbers can be arranged from 6690690?

emilem · Answer

christoph, how about 0s when they come in the first place?

Shahzad · Answer

emilem
 u r rite n this is the trick in that v need to takecare off

in my opinion answer is 60

solution:- 6!/(31*2!*2!) * 2

explanation:- first v will permutate for last 6 places were v can fit in any of the three digits i.e 6,9,0 and multiply it by two for first position where only two digits can come

christoph · Answer

!?**

you are right ! is it 150 ? 6!/(2!*2!*2!) + 6!/(3!*2!)

mirhaque · Answer

0=2times
6=3times
9=2times

first digit can't be zero, so that leaves with any of the five digits:
5*6!/2!*3!*2!

july05 · Answer

do we consider a case when there are 2 "O"s at the beginning?

7!/(3! 2! 2!) - 6!/(3! 2! 1!) -5!/(3! 2!) ????

sparky · Answer

julie

I don't think you need to include a case with 2 zeros up front becasue those cases are already included in 6!/(3! 2! 1!)

emilem · Answer

Folks, the OA is 150. The interesting method for solving this problem that I came across in a book is that first we need to find total no. of possibilities, which is 7!/(3!*2!*2!)=210. Then, assess the position of 0s. Since we have 2 zeros out of 7 digits that can stand in the first place, 2/7th of total possibilities will include 0 in the first place. Therefore, eliminating them we have 5/7th of total possibilities or 5/7*210=150.

Do you think, guys, this solution is applicable to all this kind of problems? Just by using fractions, it seems pretty easy and logical, but would like to see your comments. By the way, other solutions are really good.

How many 7-digit numbers can be arranged from 6690690

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