Permutations and Combinations | Class 11 Mathematics Notes
By ConceptScroll Team · Published on 17 July 2026 · 3 min read
Permutations and Combinations – this guide gives you a concise, exam-ready overview of Permutations and Combinations from Class 11 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
6.4 Combinations
This section introduces combinations, which are selections of objects where order does not matter. Using the example of forming a team of 2 players from 3 players X, Y, Z, it shows that the possible teams are XY, YZ, and ZX. Unlike permutations, the order of selection is irrelevant in combinations. The section explains that the number of combinations of n distinct objects taken r at a time, denoted by nCr, is related to permutations by the formula nPr = nCr × r!, since each combination corresponds to r! permutations. The formula for combinations is derived as nCr = n! / (r! (n-r)!). Properties such as nCr = nC(n-r) and the definition of nC0 = 1 are discussed. The section also presents Theorem 6: nCr + nC(r-1) = (n+1)Cr, with proof. Examples include counting handshakes, chords, committees, and card selections. The section emphasizes the difference between permutations (order matters) and combinations (order does not matter) and provides formulas and examples for calculating combinations in various contexts.
📊 Diagram: Fig. 6.3; These are XY, YZ and ZX (Fig 6.3).
🧪 Activity: No specific activity, but multiple examples illustrate the concept.
🔗 Connection: Leads to exercises on combinations and miscellaneous examples combining permutations and combinations.
Frequently asked questions
1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed?
(i) When repetition of digits is allowed, each of the 3 places can be filled with any of the 5 digits. So, number of 3-digit numbers = 5 × 5 × 5 = 125.
(ii) When repetition is not allowed, the first digit can be chosen in 5 ways, the second in 4 ways (excluding the first digit), and the third in 3 ways. So, number of 3-digit numbers = 5 × 4 × 3 = 60.
2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
To form a 3-digit even number, the last digit must be even. Digits available: 1,2,3,4,5,6 Even digits: 2,4,6 (3 choices for last digit) Since repetition is allowed: First digit: 6 choices (1 to 6) Second digit: 6 choices (1 to 6) Third digit (last digit): 3 choices (2,4,6) Total numbers = 6 × 6 × 3 = 108.
3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
First 10 letters: A, B, C, D, E, F, G, H, I, J No repetition allowed. Number of 4-letter codes = Number of permutations of 10 letters taken 4 at a time = {}^{10}P_4 = 10 × 9 × 8 × 7 = 5040.
4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
The telephone number is 5-digit starting with 6 and 7. Digits used: 0 to 9 First two digits fixed: 6 and 7 Remaining 3 digits to be chosen from digits excluding 6 and 7 (digits left = 8) No repetition allowed. Number of ways = permutations of 8 digits taken 3 at a time = {}^{8}P_3 = 8 × 7 × 6 = 336.
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