How many 5 digit numbers can be formed from 12345 if repetition is allowed?

How many 5-digit prime numbers can be formed using the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?

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• How many 5-digit prime numbers can be formed using the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?
• Answer (Detailed Solution Below)
• How many 5 digit numbers can be formed from 12345 which are divisible by 3?
• How many 5 digit numbers can be made which are divisible by 3?
• How many five digit numbers can be formed from the digits 12345?

NDA (Held On: 18 Apr 2021) Maths Previous Year paper

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1. 5
2. 4
3. 3
4. 0

Answer (Detailed Solution Below)

Option 4 : 0

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Electric charges and coulomb's law (Basic)

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Concept:

Prime number: Prime number are those which are divisible by itself and 1.

Calculation:

Not a single five-digit prime number can be formed using the digits 1, 2, 3, 4, 5(without repetition).

This is because if one adds the digits, the result obtained will be = 1 + 2 + 3 + 4 + 5 = 15 which is divisible by 3.

Hence, any number obtained as a permutation of these 5 digits will be at least divisible by 3 and cannot be a prime number.

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How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?

Five-digits numbers divisible by 3 are to be formed using the digits 0, 1, 2, 3, 4, and 5 without repetition.
For a number to be divisible by 3, the sum of its digits should be divisible by 3,
Consider the digits: 1, 2, 3, 4 and 5
Sum of the digits = 1 + 2 + 3 + 4 + 5  = 15
15 is divisible by 3.
∴ Any 5-digit number formed using the digits 1, 2, 1
Starting with the most significant digit, 5 digits are available for this place.
Since, repetition is not allowed, for the next significant place, 4 digits are available.
Similarly, all the places can be filled as:

Number of 5-digit numbers
= 4 × 4 × 3 × 2 × 1 = 96
Next, consider the digits: 0, 1, 2, 3, 4
Sum of the digits = 0 + 1 + 2 + 3 + 4 = 10 which is not divisible by 3.
∴ None of the 5-digit numbers formed using the digits 0, 1, 2, 3, and 4 will not be divisible by 3.
Further, no other selection of 5 digits (out of the given 6)
will give a 5-digit number, which is divisible by 3.
∴ Total number of 5adigit numbers divisible by 3
= 120 + 96 = 216

Assuming that we are considering only $5$ digit numbers (i.e. strings beginning with '$0$' are not counted) then the answer can be expressed simply as:

$$\frac{1}{3}\left(\binom{9}{5}5!+\binom{9}{4}4\cdot 4!\right)=9072\tag{Answer}$$

For the $5$ digit numbers to be multiples of $3$ we require their digit sum to be a multiple of $3$.

With this in mind, begin by considering the generating function for number of partitions into $5$ distinct parts with maximum part $9$. This generating function is related to the Gaussian binomial coefficient $\binom{n}{r}_q$. Specifically it is:

$$q^{\binom{5}{2}}\binom{9}{5}_q,$$

and the sum of it's $q^{3k}$ ($k=0,1,2\ldots$) coefficients counts distinct choices of $5$ digits from the set $\{1,2,\ldots,9\}$ that sum to multiples of $3$.

For each such choice a $5$ digit number has $5!$ permutations, therefore the number of $5$ digit numbers, divisible by $3$, using choices from $\{1,2,\ldots,9\}$ is:

$$5!\sum_{k}[q^{3k}]q^{15}\binom{9}{5}_q\tag{1}$$

Where we use the $[q^{3k}]$ operator to evaluate the coefficient of $q^{3k}$ in the generating function.

Similarly the generating function for number of partitions into $4$ distinct parts with maximum part $9$ is:

$$q^{\binom{4}{2}}\binom{9}{4}_q.$$

The sum of it's $q^{3k}$ coefficients count distinct choices of $4$ digits from the set $\{1,2,\ldots,9\}$ that sum to multiples of $3$. We may also use this sum to count the number of choices of $5$ digits from the set $\{0,1,2,\ldots,9\}$ that must include $0$ and who's digit sum is divisible by 3.

For each such choice a $5$ digit number has $4\cdot 4!$ permutations: '$0$' can take positions $2$ to $5$ then the remaining $4$ digits take the $4$ remaining positions in $4!$ ways. The number of $5$ digit numbers, divisible by $3$, using choices from $\{0,1,2,\ldots,9\}$ that must contain '$0$' is:

$$4\cdot 4!\sum_{k}[q^{3k}]q^{10}\binom{9}{5}_q\tag{2}$$

So the required answer is given by:

$$5!\sum_{k}[q^{3k}]q^{15}\binom{9}{5}_q+4\cdot 4!\sum_{k}[q^{3k}]q^{10}\binom{9}{5}_q\tag{3}$$

Using a technique called the "roots of unity filter" we see that $(1)$ gives us:

$$\sum_{k}[q^{3k}]q^{15}\binom{9}{5}_q=\frac{1}{3}\left((1)^{15}\binom{9}{5}_1+(e^{2i\pi/3})^{15}\binom{9}{5}_{e^{2i\pi/3}}+(e^{4i\pi/3})^{15}\binom{9}{5}_{e^{4i\pi/3}}\right).$$

Here we are using the cube roots of unity.

The only non-zero term on the right is $\binom{9}{5}_1=\binom{9}{5}$ since $\binom{9}{5}_q$ has a factor $(1-q^9)$ which is $0$ for $q=e^{2i\pi/3}$ and $q=e^{4i\pi/3}$. Hence

$$\sum_{k}[q^{3k}]q^{15}\binom{9}{5}_q=\frac{1}{3}\binom{9}{5}\tag{4}$$

Similarly the roots of unity filter for $(2)$ gives:

$$\sum_{k}[q^{3k}]q^{10}\binom{9}{4}_q=\frac{1}{3}\left((1)^{10}\binom{9}{4}_1+(e^{2i\pi/3})^{10}\binom{9}{4}_{e^{2i\pi/3}}+(e^{4i\pi/3})^{10}\binom{9}{4}_{e^{4i\pi/3}}\right),$$

How many 5 digit numbers can be formed if repetition is allowed?

How many 3