Question

how many 5 digits numbers can be formed if 2 3 5 always exist ?

Answer 1

Answer:

Step-by-step explanation:

abcde be the 5 digit number, if 2 3 and 5 are already there then we are left with 7 other digits.

Now, let's take 2 digits from those 7 digits and leave 0. So 6c2 ways.

And we arrange them in 5! Ways.

Hence 5!*6c2

Now, let's fix 0 and take 1 digit from rest 6 digits left. I.e 6c1

Hence 5!*6c1

But we also counted numbers with first digit as 0 i.e 4!*6c1

Answer) 6c2*5! + 6c1*5! — 6c1*4!

Answer 2

Answer:

Step-by-step explanation: