diff --git a/src/algorithms/string/combinations/__test__/combineWithoutRepetitions.test.js b/src/algorithms/string/combinations/__test__/combineWithoutRepetitions.test.js
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+++ b/src/algorithms/string/combinations/__test__/combineWithoutRepetitions.test.js
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+import combineWithoutRepetitions from '../combineWithoutRepetitions';
+import factorial from '../../../math/factorial/factorial';
+
+describe('combineWithoutRepetitions', () => {
+  it('should combine string without repetitions', () => {
+    expect(combineWithoutRepetitions('AB', 3)).toEqual([]);
+    expect(combineWithoutRepetitions('AB', 1)).toEqual(['A', 'B']);
+    expect(combineWithoutRepetitions('A', 1)).toEqual(['A']);
+    expect(combineWithoutRepetitions('AB', 2)).toEqual(['AB']);
+    expect(combineWithoutRepetitions('ABC', 2)).toEqual(['AB', 'AC', 'BC']);
+    expect(combineWithoutRepetitions('ABC', 3)).toEqual(['ABC']);
+    expect(combineWithoutRepetitions('ABCD', 3)).toEqual([
+      'ABC',
+      'ABD',
+      'ACD',
+      'BCD',
+    ]);
+    expect(combineWithoutRepetitions('ABCDE', 3)).toEqual([
+      'ABC',
+      'ABD',
+      'ABE',
+      'ACD',
+      'ACE',
+      'ADE',
+      'BCD',
+      'BCE',
+      'BDE',
+      'CDE',
+    ]);
+
+    const combinationOptions = 'ABCDEFGH';
+    const combinationSlotsNumber = 4;
+    const combinations = combineWithoutRepetitions(combinationOptions, combinationSlotsNumber);
+    const n = combinationOptions.length;
+    const r = combinationSlotsNumber;
+    const expectedNumberOfCombinations = factorial(n) / (factorial(r) * factorial(n - r));
+
+    expect(combinations.length).toBe(expectedNumberOfCombinations);
+  });
+});
diff --git a/src/algorithms/string/combinations/combineWithoutRepetitions.js b/src/algorithms/string/combinations/combineWithoutRepetitions.js
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+++ b/src/algorithms/string/combinations/combineWithoutRepetitions.js
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+/*
+  @see: https://stackoverflow.com/a/127898/7794070
+
+  Lets say your array of letters looks like this: "ABCDEFGH".
+  You have three indices (i, j, k) indicating which letters you
+  are going to use for the current word, You start with:
+
+  A B C D E F G H
+  ^ ^ ^
+  i j k
+
+  First you vary k, so the next step looks like that:
+
+  A B C D E F G H
+  ^ ^   ^
+  i j   k
+
+  If you reached the end you go on and vary j and then k again.
+
+  A B C D E F G H
+  ^   ^ ^
+  i   j k
+
+  A B C D E F G H
+  ^   ^   ^
+  i   j   k
+
+  Once you j reached G you start also to vary i.
+
+  A B C D E F G H
+    ^ ^ ^
+    i j k
+
+  A B C D E F G H
+    ^ ^   ^
+    i j   k
+  ...
+ */
+
+/**
+ * @param {string} combinationOptions
+ * @param {number} combinationLength
+ * @return {string[]}
+ */
+export default function combineWithoutRepetitions(combinationOptions, combinationLength) {
+  // If combination length is just 1 then return combinationOptions.
+  if (combinationLength === 1) {
+    return Array.from(combinationOptions);
+  }
+
+  // Init combinations array.
+  const combinations = [];
+
+  for (let i = 0; i <= (combinationOptions.length - combinationLength); i += 1) {
+    const smallerCombinations = combineWithoutRepetitions(
+      combinationOptions.substr(i + 1),
+      combinationLength - 1,
+    );
+
+    for (let j = 0; j < smallerCombinations.length; j += 1) {
+      combinations.push(combinationOptions[i] + smallerCombinations[j]);
+    }
+  }
+
+  // Return all calculated combinations.
+  return combinations;
+}