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Permutations and Combinations - Permutations when all the objects are not distinct

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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When objects are not distinct, the standard n!n! formula for permutations counts identical arrangements as different. To correct this, we divide by the factorial of the count of each set of identical objects.

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If we have nn objects where pp objects are of one kind (identical) and the rest are all distinct, the number of permutations is n!p!\frac{n!}{p!}.

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In a more general case, if there are nn objects where p1p_1 are of one kind, p2p_2 are of a second kind, ..., pkp_k are of a kthk^{th} kind, such that p1+p2+⋯+pk=np_1 + p_2 + \dots + p_k = n, the number of unique permutations is given by the multinomial coefficient.

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This concept is frequently applied to problems involving the rearrangement of letters in words where certain letters repeat (e.g., 'ROOT', 'INSTITUTE', 'MISSISSIPPI').

📐Formulae

Permutations of n objects with p identical objects=n!p!\text{Permutations of } n \text{ objects with } p \text{ identical objects} = \frac{n!}{p!}

Total Permutations=n!p1!p2!…pk!\text{Total Permutations} = \frac{n!}{p_1! p_2! \dots p_k!}

💡Examples

Problem 1:

Find the number of ways to rearrange the letters of the word APPLEAPPLE.

Solution:

In the word APPLEAPPLE, there are n=5n = 5 letters in total. The letter PP appears 22 times, and the letters AA, LL, and EE appear 11 time each. Using the formula for non-distinct objects: Number of ways=5!2!=1202=60\text{Number of ways} = \frac{5!}{2!} = \frac{120}{2} = 60

Explanation:

We divide by 2!2! because the two PP's are identical, and swapping them does not create a new unique arrangement.

Problem 2:

How many different signals can be generated by arranging 66 flags in a line, if 22 are red, 33 are yellow and 11 is blue?

Solution:

Total number of flags n=6n = 6. Identical flags are: Red (p1=2p_1 = 2), Yellow (p2=3p_2 = 3), and Blue (p3=1p_3 = 1). Total signals=6!2!×3!×1!\text{Total signals} = \frac{6!}{2! \times 3! \times 1!} Total signals=7202×6×1=72012=60\text{Total signals} = \frac{720}{2 \times 6 \times 1} = \frac{720}{12} = 60

Explanation:

The total permutations are divided by the factorials of the counts of each color to account for the fact that flags of the same color are indistinguishable.

Problem 3:

In how many ways can the letters of the word MATHEMATICSMATHEMATICS be arranged?

Solution:

The word MATHEMATICSMATHEMATICS has 1111 letters. The frequencies are: M=2M = 2, A=2A = 2, T=2T = 2, H=1H = 1, E=1E = 1, I=1I = 1, C=1C = 1, S=1S = 1. Total arrangements=11!2!⋅2!⋅2!\text{Total arrangements} = \frac{11!}{2! \cdot 2! \cdot 2!} Total arrangements=399168002×2×2=399168008=4989600\text{Total arrangements} = \frac{39916800}{2 \times 2 \times 2} = \frac{39916800}{8} = 4989600

Explanation:

Since MM, AA, and TT each repeat twice, we divide the total 11!11! by (2!)3(2!)^3.