Question

How many numbers which are
1.Even formed by taking all the digits 1, 2, 3, 4, 5?
2. Less than 40,000 can be formed by taking all the digits 1, 2, 3, 4, 5?
3. The number of odd numbers between 1000 and 10,000 can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 is?

Answer 1

Hint: In this problem, we have to find how many possible digits less than 40,000 can be formed by taking all the digits 1, 2, 3, 4, 5 and the number of odd numbers between 1000 and 10,000 can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. Here we can first find the even terms with the given digits and we can find the following answers in the solution part.

Complete step by step answer:
1.Here we can take the numbers. 1, 2, 3, 4, 5.
We know that, even numbers which will have 2 in the last and remaining four numbers are replaced to get possible 5 digit even numbers. We will now have the factorial of 4, where 4 digits are replaced continuously.
\[\Rightarrow 4!=24\]
Similarly, in that 5-digit number, we have 4 as the even number at last, then we will get
\[\Rightarrow 4!=24\]
Hence, even numbers formed from 1, 2, 3, 4, 5 are,
\[\Rightarrow 24+24=48\]
2. We are given 1, 2, 3, 4, 5 from which we have to find the number less than 40,000.
We can now see that 40,000 is a 5 digit number, where the numbers less than 40,000 will have 1 or 2 or 3 in the first place, so similar to problem, 1, we can write it as,
\[\Rightarrow 24+24+24=72\]
3. here we have to find the number of odd numbers between 1000 and 10,000 that can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9.
We can see that the numbers between 1000 and 10,000 will be a 4-digit number.
We have 8 numbers given.
As the unit place is an odd, it can be filled in 5 ways by any of the 5 odd numbers. Out of the remaining 8 ways we have to arrange 3 ways.
\[{{\Rightarrow }^{8}}{{P}_{3}}=\dfrac{8!}{5!}=8\times 7\times 6=336\]
We can now multiply the 5 ways, we get
\[\Rightarrow 336\times 5=1680\]

Note: We should always remember that the formula of permutation is \[^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\] where n is the number of terms and r is the order in which it is arranged. We should also know that, for example if n terms are replaced with each other, then we can write it as the factorial as \[n!\].