1. ## Permutation (URGENT)

Hello all. I seriously need an urgent help from y'all.If you would so kind , could anyone of you please solve these two questions , which is of permutation. My brain couldn't go further :

"How many numbers greater than 5000 can be formed using he digits 3 , 4 , 5 , 6 and 7 without repetition of digits? How many of these are even numbers?"

AND

" How many numbers between 2000 and 8999 (both inclusive) do not have four different digits ? "

please save me, im almost drowned of solving this kind of things

2. Hello, discovery!

Here's the second one . . .

How many numbers between 2000 and 8999 (both inclusive)
do not have four different digits ?

There are $7000$ numbers from 2000 to 8999.

We'll find the number of numbers that do have four different digits.
. . The first digit has 7 choices: $\{2,3,4,5,6,7,8\}$
. . The other three digits are any permutation of the remaining 9 digits: . $P(9,3) \:=\:504$
. . Hence, there are: . $7\cdot504 \:=\:3528$ numbers with four different digits.

Therefore, there are: . $7000 - 3528 \:=\:3472$ numbers that do not have four different digits.

3. For the first one (I hope I don't miss some).

The number has to be > 5000. for the 4 digit numbers the first digit must be a 5, 6 or 7. We place the 5 in the first spot and arrange 3 of the remaining 4
digits in P(4,3)=24 ways.

Do the same for 6 and 7.

Then for the remaining 5 digit numbers we have 5!=120 ways.

There are 120+24+24+24=192 numbers.

4. Thanksssssss!!! I Got It !!

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# how many numbers between 2000 and 8999 (both inclusive) do not have four different digits

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