Solving Some Basic Questions on Permutation & Combination

Duration: 14 min

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This educational video is a lecture on permutations and combinations, presented by an instructor from Knowledge Gate Educator. The video begins with an introduction to the topic, using visual metaphors of a roulette wheel and a fruit basket to illustrate the difference between permutations (order matters) and combinations (order does not matter). The core of the lecture consists of several worked examples. The first problem asks for the number of ways to seat 10 people in 4 chairs, which is solved using the permutation formula P(n,r) = n! / (n-r)!, resulting in 10P4 = 10 × 9 × 8 × 7 = 5040. The second example calculates the total number of 5-lettered words from the 26 alphabets, first without repetition (26P5 = 26 × 25 × 24 × 23 × 22) and then with repetition allowed (26^5). The third problem involves finding the number of 3-digit numbers between 100 and 999 that do not contain the digit 7, which is solved by considering the available digits for each place value (8 choices for the hundreds, 9 for tens, and 9 for units, resulting in 8 × 9 × 9 = 648). The final example asks for the number of 5-lettered words that start with 'a' and end with 'z' without repetition, which is calculated as 24P3 = 24 × 23 × 22 = 12144. The video concludes with a 'Thanks for Watching' screen.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a title card for a lesson on 'PERMUTATION & COMBINATION'. The title is split into two parts, with 'PERMUTATION' over a roulette table and dice, and 'COMBINATION' over a fruit basket and a pizza. This visual metaphor illustrates the core concept: permutations are about ordered arrangements (like a lock combination), while combinations are about unordered selections (like a fruit basket). The scene then transitions to a presentation slide titled 'Confused' with the question 'Should I unlock with Permutation or Combination?'. This slide features a cartoon detective with a magnifying glass, standing between two locked doors labeled 'Permutation' and 'Combination', visually representing the confusion students often feel. The instructor, Yash Jain Sir, is shown in a small window, and the video title 'APTITUDE PROBLEMS ON PERMUTATION & COMBINATION' is displayed on a large yellow background.

  2. 2:00 – 5:00 02:00-05:00

    The instructor presents the first problem: 'If we have 10 people and 4 chairs, what are the number of ways to seat them?'. The on-screen text clearly states the question. The instructor explains that this is a permutation problem because the order of seating matters. He writes the permutation formula P(n,r) = n! / (n-r)! and substitutes n=10 and r=4, writing 10P4. He then expands the formula as 10! / (10-4)! = 10! / 6!. He further simplifies this by writing out the factorial as 10 × 9 × 8 × 7 × 6! / 6!, and the 6! terms cancel out, leaving the final calculation of 10 × 9 × 8 × 7.

  3. 5:00 – 10:00 05:00-10:00

    The second problem is introduced: 'Find the total number of 5 lettered words??'. The instructor presents two sub-questions: a) No repetition allowed, and b) Repetition allowed. For part (a), he explains that there are 26 alphabets (A-Z) and 5 slots. He writes the calculation as 26 × 25 × 24 × 23 × 22, which is 26P5. For part (b), he explains that with repetition allowed, each of the 5 slots can be filled with any of the 26 letters, so the total is 26^5. The instructor then moves to the third problem: 'Number of numbers between 100-999 which does not contain 7?'. He explains that this is a 3-digit number problem. He identifies the available digits for each place: for the hundreds place, digits 1-9 are available, but 7 is excluded, so 8 choices (1,2,3,4,5,6,8,9). For the tens and units places, digits 0-9 are available, but 7 is excluded, so 9 choices each (0,1,2,3,4,5,6,8,9). He writes the calculation as 8 × 9 × 9.

  4. 10:00 – 13:54 10:00-13:54

    The final problem is presented: 'In how many ways can we form a 5 lettered word? No repetition, starting with 'a' and ending with 'z'.'. The instructor explains that the first and last positions are fixed. The first letter is 'a' and the last is 'z', so the middle three positions are the only ones that can vary. He states that there are 24 letters left to choose from (26 total minus 'a' and 'z'). He then calculates the number of ways to fill the 3 middle slots as 24P3, which is 24 × 23 × 22. He performs the multiplication to get the final answer of 12144. The video concludes with a black screen displaying the text 'THANKS FOR WATCHING' in an orange and white box.

The video provides a structured and practical introduction to permutations and combinations. It begins by establishing the fundamental difference between the two concepts using relatable visual analogies. The core of the lesson is a series of progressively challenging, real-world problems. The instructor methodically applies the permutation formula P(n,r) to problems involving seating arrangements and word formation, clearly demonstrating the step-by-step calculation. He also addresses the case of repetition allowed, which is a key variation. The final problem integrates multiple constraints (fixed start and end letters), reinforcing the application of the concepts. The overall teaching style is clear, with a focus on problem-solving methodology, making it a useful resource for students preparing for aptitude tests.