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Question

Question: How many different numbers can be formed with the digits 1, 3, 5, 7 and 9? A) 325 B) 576 C) 62...

How many different numbers can be formed with the digits 1, 3, 5, 7 and 9?
A) 325
B) 576
C) 625
D) 525

Explanation

Solution

We will find the number of possible 1 digit numbers, 2 digit numbers, 3 digit numbers, 4 digit numbers and as well as 5 digit numbers using the given five digits. Then, we will add these numbers of digits possible to form to get the required answer.

Complete step-by-step answer:
We can form any digit number: 1, 2, 3, 4 or 5. We will find every possibility.
If we need to make one digit number, we can basically get any number among the given digits only which are 1, 3, 5, 7 and 9. So, we can therefore form 5 digits.
Hence, one digits numbers = 5. ……………….(1)
Now, let come over two digit numbers:
To form 2 digit numbers, we need to choose 2 numbers out of 5 and then permute them as well.
Therefore, the possible two digit numbers are = 5C2×2^5{C_2} \times 2
Now, we know that we have the following formula:-
nCr=n!r!(nr)!{ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}} …………..(2)
Therefore, the possible two digit numbers are = 5!3!×2!×2\dfrac{{5!}}{{3! \times 2!}} \times 2
On simplifying it by using n!=n.(n1)......1n! = n.(n - 1)......1 …………(3), we can write the above expression as:
\therefore Possible two digit numbers are = 5×5×2=505 \times 5 \times 2 = 50 …………..(4)
Now, let come over three digit numbers:
To form 3 digit numbers, we need to choose 3 numbers out of 5 and then permute them as well.
Therefore, the possible three digit numbers are = 5C3×3!^5{C_3} \times 3!
Now, we will use the formula in equation (2).
Therefore, the possible three digit numbers are = 5!3!×2!×3!\dfrac{{5!}}{{3! \times 2!}} \times 3!
Now, we will use the formula in equation (3).
\therefore Possible three digit numbers are = 5×4×3=605 \times 4 \times 3 = 60 …………..(5)
Now, let come over four digit numbers:
To form 4 digit numbers, we need to choose 4 numbers out of 5 and then permute them as well.
Therefore, the possible four digit numbers are = 5C4×4!^5{C_4} \times 4!
Now, we will use the formula in equation (2).
Therefore, the possible four digit numbers are = 5!4!×1!×4!\dfrac{{5!}}{{4! \times 1!}} \times 4!
Now, we will use the formula in equation (3).
\therefore Possible four digit numbers are = 5×4×3×2×1=1205 \times 4 \times 3 \times 2 \times 1 = 120 …………..(6)
Now, let come over five digit numbers:
To form 5 digit numbers, we need to choose 5 numbers out of 5 and then permute them as well.
Therefore, the possible five digit numbers are = 5C5×5!^5{C_5} \times 5!
Now, we will use the formula in equation (2).
Therefore, the possible five digit numbers are = 5!5!×0!×5!\dfrac{{5!}}{{5! \times 0!}} \times 5!
Now, we will use the formula in equation (3).
\therefore Possible five digit numbers are = 5×4×3×2×1=1205 \times 4 \times 3 \times 2 \times 1 = 120 …………..(7)
Now, adding (1), (4), (5), (6) and (7), we will get:-
\Rightarrow The possible numbers are = 5 + 20 + 60 + 120 + 120 = 325.

\therefore The correct option is (A).

Note: Let us do an alternate way to solve the same question:-
If we need to make one digit number, we can basically get any number among the given digits only which are 1, 3, 5, 7 and 9. So, we can therefore form 5 digits.
Hence, one digits numbers = 5. ……………….(A)
Now, let come over two digit numbers:
To form 2 digit numbers, we need to fill __ __ with 5 given digits.
Since, the first place can be filled with 5 digits and second place by 4 digits.
\therefore Number of 2 digit numbers = 5×4=205 \times 4 = 20 …………….(B)
Now, let come over three digit numbers:
To form 3 digit numbers, we need to fill __ __ __ with 5 given digits.
Since, the first place can be filled with 5 digits, second place by 4 digits and third place by 3 digits.
\therefore Number of 3 digit numbers = 5×4×3=605 \times 4 \times 3 = 60 …………….(C)
Now, let come over four digit numbers:
To form 4 digit numbers, we need to fill __ __ __ __ with 5 given digits.
Since, the first place can be filled with 5 digits, second place by 4 digits, third place by 3 digits and fourth place by 2 digits.
\therefore Number of 4 digit numbers = 5×4×3×2=1205 \times 4 \times 3 \times 2 = 120 …………….(D)
Now, let come over five digit numbers:
To form 4 digit numbers, we need to fill __ __ __ __ __ with 5 given digits.
Since, the first place can be filled with 5 digits, second place by 4 digits, third place by 3 digits, fourth place by 2 digits and fifth place by 1 digit.
\therefore Number of 5 digit numbers = 5×4×3×2×1=1205 \times 4 \times 3 \times 2 \times 1 = 120 …………….(E)
Now, adding (A), (B), (C), (D) and (E), we will get:-
\Rightarrow The possible numbers are = 5 + 20 + 60 + 120 + 120 = 325.
\therefore the correct option is (A).