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# N digit numbers divisible by 5 formed from the M digits

• Last Updated : 11 May, 2021

Given M unique digits and a number N. The task is to find the number of N-digit numbers which can be formed from the given M digits, which are divisible by 5 and none of the digits is repeated.
Note: If it is not possible to form a N digit number from the given digits, print -1.
Examples:

Input : N = 3, M = 6, digits[] = {2, 3, 5, 6, 7, 9}
Output : 20
Input : N = 5, M = 6, digits[] = {0, 3, 5, 6, 7, 9}
Output : 240

For a number to be divisible by 5, the only condition is that the digit at the unit place in the number must be either 0 or 5.
So, to find the count of numbers that are divisible by 5 and can be formed from the given digits, do the following:

• Check if the given digits contain both 0 and 5.
• If the given digits contain both 0 and 5, then the unit place can be filled in 2 ways otherwise the unit place can be filled in 1 way.
• Now, the tens place can now be filled by any of the remaining M-1 digits. So, there are (M-1) ways of filling the tens place.
• Similarly, the hundred’s place can now be filled by any of the remaining (M-2) digits and so on.

Therefore, if the given digits have both 0 and 5:

`Required number of numbers = 2 * (M-1)* (M-2)...N-times. `

Otherwise if the given digits have either one of 0 and 5 and not both:

`Required number of numbers = 1 * (M-1)* (M-2)...N-times. `

Below is the implementation of the above approach.

## C++

 `// CPP program to find the count of all``// possible N digit numbers which are``// divisible by 5 formed from M digits``#include ``using` `namespace` `std;` `// Function to find the count of all``// possible N digit numbers which are``// divisible by 5 formed from M digits``int` `numbers(``int` `n, ``int` `arr[], ``int` `m)``{``    ``int` `isZero = 0, isFive = 0;``    ``int` `result = 0;` `    ``// If it is not possible to form``    ``// n digit number from the given``    ``// m digits without repetition``    ``if` `(m < n) {``        ``return` `-1;``    ``}` `    ``for` `(``int` `i = 0; i < m; i++) {``        ``if` `(arr[i] == 0)``            ``isZero = 1;` `        ``if` `(arr[i] == 5)``            ``isFive = 1;``    ``}` `    ``// If both zero and five exists``    ``if` `(isZero && isFive) {``        ``result = 2;` `        ``// Remaining N-1 iterations``        ``for` `(``int` `i = 0; i < n - 1; i++) {``            ``result = result * (--m);``        ``}``    ``}``    ``else` `if` `(isZero || isFive) {``        ``result = 1;` `        ``// Remaining N-1 iterations``        ``for` `(``int` `i = 0; i < n - 1; i++) {``            ``result = result * (--m);``        ``}``    ``}``    ``else``        ``result = -1;` `    ``return` `result;``}` `// Driver code``int` `main()``{``    ``int` `n = 3, m = 6;` `    ``int` `arr[] = { 2, 3, 5, 6, 7, 9 };` `    ``cout << numbers(n, arr, m);` `    ``return` `0;``}`

## Java

 `// Java program to find the count of all``// possible N digit numbers which are``// divisible by 5 formed from M digits` `class` `GFG {` `// Function to find the count of all``// possible N digit numbers which are``// divisible by 5 formed from M digits``    ``static` `int` `numbers(``int` `n, ``int` `arr[], ``int` `m) {``        ``int` `isZero = ``0``, isFive = ``0``;``        ``int` `result = ``0``;` `        ``// If it is not possible to form``        ``// n digit number from the given``        ``// m digits without repetition``        ``if` `(m < n) {``            ``return` `-``1``;``        ``}` `        ``for` `(``int` `i = ``0``; i < m; i++) {``            ``if` `(arr[i] == ``0``) {``                ``isZero = ``1``;``            ``}` `            ``if` `(arr[i] == ``5``) {``                ``isFive = ``1``;``            ``}``        ``}` `        ``// If both zero and five exists``        ``if` `(isZero == ``1` `&& isFive == ``1``) {``            ``result = ``2``;` `            ``// Remaining N-1 iterations``            ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {``                ``result = result * (--m);``            ``}``        ``} ``else` `if` `(isZero == ``1` `|| isFive == ``1``) {``            ``result = ``1``;` `            ``// Remaining N-1 iterations``            ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {``                ``result = result * (--m);``            ``}``        ``} ``else` `{``            ``result = -``1``;``        ``}` `        ``return` `result;``    ``}` `// Driver code``    ``public` `static` `void` `main(String[] args) {``        ``int` `n = ``3``, m = ``6``;` `        ``int` `arr[] = {``2``, ``3``, ``5``, ``6``, ``7``, ``9``};``        ``System.out.println(numbers(n, arr, m));` `    ``}``}``// This code is contributed by RAJPUT-JI`

## Python 3

 `# Python 3 program to find the count``# of all possible N digit numbers which``# are divisible by 5 formed from M digits` `# Function to find the count of all``# possible N digit numbers which are``# divisible by 5 formed from M digits``def` `numbers(n, arr, m):` `    ``isZero ``=` `0``    ``isFive ``=` `0``    ``result ``=` `0` `    ``# If it is not possible to form``    ``# n digit number from the given``    ``# m digits without repetition``    ``if` `(m < n) :``        ``return` `-``1` `    ``for` `i ``in` `range``(m) :``        ``if` `(arr[i] ``=``=` `0``):``            ``isZero ``=` `1` `        ``if` `(arr[i] ``=``=` `5``):``            ``isFive ``=` `1` `    ``# If both zero and five exists``    ``if` `(isZero ``and` `isFive) :``        ``result ``=` `2` `        ``# Remaining N-1 iterations``        ``for` `i ``in` `range``( n ``-` `1``):``            ``m ``-``=` `1``            ``result ``=` `result ``*` `(m)` `    ``elif` `(isZero ``or` `isFive) :``        ``result ``=` `1` `        ``# Remaining N-1 iterations``        ``for` `i ``in` `range``(n ``-` `1``) :``            ``m ``-``=` `1``            ``result ``=` `result ``*` `(m)``    ``else``:``        ``result ``=` `-``1` `    ``return` `result` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ``n ``=` `3``    ``m ``=` `6` `    ``arr ``=` `[ ``2``, ``3``, ``5``, ``6``, ``7``, ``9``]` `    ``print``(numbers(n, arr, m))` `# This code is contributed by ChitraNayal`

## C#

 `// C# program to find the count of all``// possible N digit numbers which are``// divisible by 5 formed from M digits``using` `System;``public` `class` `GFG {` `// Function to find the count of all``// possible N digit numbers which are``// divisible by 5 formed from M digits``    ``static` `int` `numbers(``int` `n, ``int` `[]arr, ``int` `m) {``        ``int` `isZero = 0, isFive = 0;``        ``int` `result = 0;` `        ``// If it is not possible to form``        ``// n digit number from the given``        ``// m digits without repetition``        ``if` `(m < n) {``            ``return` `-1;``        ``}` `        ``for` `(``int` `i = 0; i < m; i++) {``            ``if` `(arr[i] == 0) {``                ``isZero = 1;``            ``}` `            ``if` `(arr[i] == 5) {``                ``isFive = 1;``            ``}``        ``}` `        ``// If both zero and five exists``        ``if` `(isZero == 1 && isFive == 1) {``            ``result = 2;` `            ``// Remaining N-1 iterations``            ``for` `(``int` `i = 0; i < n - 1; i++) {``                ``result = result * (--m);``            ``}``        ``} ``else` `if` `(isZero == 1 || isFive == 1) {``            ``result = 1;` `            ``// Remaining N-1 iterations``            ``for` `(``int` `i = 0; i < n - 1; i++) {``                ``result = result * (--m);``            ``}``        ``} ``else` `{``            ``result = -1;``        ``}` `        ``return` `result;``    ``}` `// Driver code``    ``public` `static` `void` `Main() {``        ``int` `n = 3, m = 6;` `        ``int` `[]arr = {2, 3, 5, 6, 7, 9};``        ``Console.WriteLine(numbers(n, arr, m));` `    ``}``}``// This code is contributed by RAJPUT-JI`

## PHP

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## Javascript

 ``
Output:
`20`

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