Permutations Question Help!

Question

1.How many numbers greater than 300,000 are there using only the digits 1,1,1,2,2,3?

How would do this question, please explain with additional steps.

Answer 1

In this problem, we can't use 1 or 2 as the hundred thousandth digit, or else it will become less than 300,000. So we have to fix 3 as the beginning digit. Therefore only the remaining 5 digits are subject for arrangement (three 1's and and two 2's). We use permutation of repeated items in this case:

5!/(3!2!)= 10 numbers.

Answer 2

For the number to be greater than 300,000 , the 3 must be in the hundred thousands place. After that, we have 5 digits to permute, but 3 are repeated and 2 are repeated. So we have 5! / (3!·2!) = 10. So few, in fact, we could probably list them all:

a) 311,122 b) 311,212 c) 311,221 d) 312,112 e) 312,121 f) 312,211 g) 321,112 h) 321,121 i) 321,211 j) 322,111