Permutations and Combinations Ex 7.3: Class 11 Maths NCERT Solutions

Understanding the Permutation Formula: nPr

A permutation is an arrangement of objects in a specific order. When we want to arrange 'r' objects taken from a total of 'n' distinct objects, we use the permutation formula. This is denoted as P(n, r) or, more commonly, nPr.

The formula is:
nPr = n! / (n-r)!

Where:

• n is the total number of distinct objects available.
• r is the number of objects we are selecting and arranging.
• n! (n factorial) is the product of all positive integers up to n (e.g., 5! = 5 × 4 × 3 × 2 × 1).

Why does this formula work? Imagine you have 'n' items and 'r' empty slots to fill. For the first slot, you have 'n' choices. For the second, you have 'n-1' choices, and so on, until the r-th slot, for which you have 'n-r+1' choices. By the multiplication principle, the total number of arrangements is n × (n-1) × ... × (n-r+1). This can be written more compactly using factorials as n! / (n-r)!. We divide by (n-r)! to cancel out the terms from (n-r) down to 1, which represents the items we did not choose to arrange.

Permutations with Non-Distinct Objects

What if some of the objects you are arranging are identical? For example, how many distinct ways can you arrange the letters of the word 'ROOT'? If the two 'O's were distinct (say, O1 and O2), we'd have 4! = 24 arrangements. But RO1O2T and RO2O1T look the same: 'ROOT'. We've overcounted.

To correct this, we use a modified formula. The number of distinct permutations of 'n' objects, where there are p1 objects of one kind, p2 objects of a second kind, ..., and pk objects of a k-th kind, is given by:

**n! / (p1! p2! ... * pk!)**

For the word 'ROOT':

• n = 4 (total letters)
• The letter 'O' is repeated 2 times (p1 = 2).
• The number of arrangements is 4! / 2! = (4 × 3 × 2 × 1) / (2 × 1) = 12.

We divide by 2! because the two 'O's can be arranged among themselves in 2! ways, and all those arrangements are indistinguishable.

Worked Examples for Ex 7.3 Concepts

• Example 1: Using the nPr Formula How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5 assuming that repetition of the digits is not allowed? Step 1: Identify n and r. We have a total of 5 distinct digits to choose from, so n = 5. We need to form 3-digit numbers, so we are arranging 3 digits at a time. Thus, r = 3. Step 2: Apply the permutation formula nPr = n! / (n-r)!. We need to calculate 5P3. 5P3 = 5! / (5-3)! = 5! / 2! Step 3: Calculate the factorial values and solve. 5! = 5 × 4 × 3 × 2 × 1 = 120 2! = 2 × 1 = 2 5P3 = 120 / 2 = 60. Final Answer: There are 60 different 3-digit numbers that can be formed.
• Example 2: Permutations with Non-Distinct Objects Find the number of different 8-letter arrangements that can be made from the letters of the word 'DAUGHTER' so that all vowels occur together. Step 1: Group the objects that must be together. The vowels in 'DAUGHTER' are A, U, E. Let's treat this group (AUE) as a single object or block. Step 2: Find the number of objects to arrange now. We now have the consonants D, G, H, T, R and the single block of vowels (AUE). This gives us a total of 5 + 1 = 6 objects to arrange. Step 3: Calculate the permutations of these new objects. These 6 distinct objects can be arranged in 6! ways. 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720. Step 4: Calculate the internal permutations of the grouped object. The vowels within the block (A, U, E) are distinct and can be arranged among themselves in 3! ways. 3! = 3 × 2 × 1 = 6. Step 5: Apply the multiplication principle. For each of the 720 arrangements of the blocks, there are 6 internal arrangements of the vowels. Total arrangements = (Arrangements of blocks) × (Internal arrangements of vowels) Total arrangements = 720 × 6 = 4320. Final Answer: There are 4320 arrangements where all vowels occur together.

Exam Traps and Key Pointers for Ex 7.3

Students often make a few common mistakes in this section. Be careful!

1. Confusing the Two Formulas: The most common error is using the simple nPr formula when objects are repeated, or using the division formula (n! / p!) when all objects are distinct. Always ask yourself: Are all the items distinct? If yes, use nPr. If no (like letters in 'APPLE'), use the division formula.

1. Incorrectly Identifying 'n' and 'r': Read the question carefully. 'n' is the total number of items available, while 'r' is the number you are choosing and arranging. In word problems with constraints (like 'vowels together'), 'n' changes after you group the items.

1. Forgetting to Divide for ALL Repeated Items: In a word like 'MISSISSIPPI', you must divide by the factorials for all repeated letters: 4! for 'I', 4! for 'S', and 2! for 'P'. Forgetting one of these will lead to an incorrect answer.

Practice Questions with Solutions

• Q: Find the value of n such that nP5 = 42 × nP3, for n > 4. A: Step 1: Write out the permutation formulas for both sides of the equation. nP5 = n! / (n-5)! nP3 = n! / (n-3)! Step 2: Substitute these into the given equation. n! / (n-5)! = 42 × [n! / (n-3)!] Step 3: Simplify the equation. We can cancel n! from both sides (since n > 4, n! is not zero). 1 / (n-5)! = 42 / (n-3)! Step 4: Expand the larger factorial to relate it to the smaller one. (n-3)! = (n-3)(n-4)(n-5)! So, 1 / (n-5)! = 42 / [(n-3)(n-4)(n-5)!] Step 5: Cancel (n-5)! and solve the resulting quadratic equation. 1 = 42 / [(n-3)(n-4)] (n-3)(n-4) = 42 n² - 7n + 12 = 42 n² - 7n - 30 = 0 (n-10)(n+3) = 0 This gives n = 10 or n = -3. Since n must be a positive integer, we discard n = -3. Final answer: n = 10.
• Q: How many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together? A: Step 1: First, find the total number of distinct permutations of the word MISSISSIPPI. Total letters (n) = 11. Repeated letters: M=1, I=4, S=4, P=2. Total permutations = 11! / (4! 4! 2!) = (39916800) / (24 24 2) = 34650. Step 2: Now, find the number of permutations where the four I's do come together. Treat the four I's (IIII) as a single block. Now we have 11 - 4 + 1 = 8 objects to arrange: (IIII), M, S, S, S, S, P, P. Number of such arrangements = 8! / (4! 2!) = 40320 / (24 2) = 840. Step 3: The number of permutations where the I's do not come together is (Total permutations) - (Permutations where they do come together). Number of required permutations = 34650 - 840 = 33810. Final answer: 33810.
• Q: Find the number of words, with or without meaning, which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the 50th word? A: Step 1: Find the number of permutations of AGAIN. 'AGAIN' has 5 letters, with 'A' repeated twice. Total words = 5! / 2! = 120 / 2 = 60. Step 2: To find the 50th word, list words in alphabetical order (A, G, I, N). Words starting with A: The remaining letters are G, I, N, A. These can be arranged in 4! = 24 ways. Words starting with G: The remaining letters are A, A, I, N. These can be arranged in 4! / 2! = 12 ways. Words starting with I: The remaining letters are A, A, G, N. These can be arranged in 4! / 2! = 12 ways. Step 3: Count the words. Words starting with A, G, and I sum up to 24 + 12 + 12 = 48 words. The 49th word must start with N. The remaining letters are A, A, G, I. The first word in dictionary order is NAAGI. Step 4: The next word (the 50th) is found by arranging A, A, G, I in the next alphabetical sequence after AAGI. This would be NAAIG. Final answer: The 50th word is NAAIG.
• Q: How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed? A: Step 1: Understand the constraints. The numbers must be between 100 and 1000, which means they must be 3-digit numbers. Step 2: Consider the restrictions on each digit's place. The hundred's place cannot be 0, otherwise it would be a 2-digit number. So, there are 5 choices for the hundred's place (1, 2, 3, 4, 5). Step 3: Consider the remaining places. After filling the hundred's place, we have 5 digits left (including 0). We need to fill the ten's and unit's places from these 5 remaining digits. This is an arrangement of 2 digits from 5 available digits, which is 5P2. Number of ways to fill the last two places = 5P2 = 5! / (5-2)! = 5! / 3! = 5 × 4 = 20. Step 4: Apply the multiplication principle. The total number of ways is the product of the number of choices for each step. Total numbers = (Choices for hundred's place) × (Ways to fill remaining two places) = 5 × 20 = 100. Final answer: 100.

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