How many distinct positive integers are possible with the digits 1, 3, 5, 7 without repetition?

Hint: First, we have to identify whether this problem is related to permutation or combination. As per the question, we have to make possible integers without any repetition of same number of this will be solved by using permutation given as ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ . So, order of the number will matter here.
Complete step-by-step answer:
Now, we have 4 digits i.e. 1, 3, 5, 7. Here we have to make all the possible numbers and then at last summation of all the numbers to see how many numbers are formed with these 4 digits.
So, first take all the one-digit numbers i.e. 1, 3, 5, 7 which will be 4 distinct numbers.
For 2-digit numbers, we will use here permutation formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ . Here n will be the total number of digits we have and r will be how many digit numbers we have to make. So, we will get
${}^{4}{{P}_{2}}=\dfrac{4!}{\left( 4-2 \right)!}$
$=\dfrac{4!}{2!}=\dfrac{4\times 3\times 2\times 1}{2\times 1}$
$=12$
So, 12 different 2-digit numbers are possible.
Similarly, now for 3-digit numbers,
${}^{4}{{P}_{3}}=\dfrac{4!}{\left( 4-3 \right)!}$
$=\dfrac{4!}{1!}=4\times 3\times 2\times 1$
$=24$
So, 24 different 3-digit numbers can be formed.
Now, for making 4-digit number we can simply use 4 factorial which can also be written as,
${}^{4}{{P}_{4}}=\dfrac{4!}{\left( 4-4 \right)!}$
$=\dfrac{4!}{0!}=\dfrac{4!}{1}$ (0! Is equal to 1)
$=4\times 3\times 2\times 1=24$
Thus, 24 different numbers can be formed for 4-digit.
Now, adding all the possible ways of 1-digit number,2-digit number, 3-digit numbers, 4-digit numbers to find total how many numbers can be formed. So,
Total integers $=$ 4 + 12 + 24 + 24
$=$ 64 integers.
Thus, a total 64 positive integers are formed without repetition.

Note: Another approach can be simple compared to this doing permutation i.e. 1-digit number can be easily identified. Talking for 2-digit numbers i.e. we have to use 4 digits which are given in question so, in the tens place there are 4 possibilities that any number can come and in one place only 3 choices are there because we don’t have to repeat the numbers. So, $4\times 3=12$ possible numbers are formed. Similarly, for 3-digit, 4-digit can be done.