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arindameducationusc

  • one year ago

Permutation Examples (Tutorial) and Basics in counting

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  1. arindameducationusc
    • one year ago
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    \(\Huge\color{red}{Fundamental~rule~of} \) \(\Huge\color{Blue}{Multiplication} \) If we take two identical jobs in which one can be completed in 'm' no. of ways & other can be completed in 'n' no. of ways. And the tasks is completed when both the jobs are completed then the total no. f way completing the task is given by \[m*n\] \(\Huge\color{Red}{Fundamental~rule~of} \) \(\Huge\color{Blue}{Addition} \) If we have two jobs in which one can be completed in 'm' no. of ways & the other can be completed in 'n' no. of ways. And the task is completed when any one of the jobs is completed then the total no. of way of completing the task is given by \[m+n\]

  2. arindameducationusc
    • one year ago
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    \(\Large\color{red}{Q} \) In a class there are 10 boys & 8 girls. A teacher wants to select a boy \(\Large\color{red}{and} \) a girl to represent the class in a function. In how many ways the selection can be done? \(\Large\color{Blue}{Ans} \) Total no. of way = 10*8=80

  3. arindameducationusc
    • one year ago
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    \(\Large\color{red}{Q} \) In a class there are 10 boys & 8 girls. A teacher wants to select a boy \(\Large\color{red}{Or} \) a girl to represent the class in a function. In how many ways the selection can be done? \(\Large\color{Blue}{Ans} \) Total no. of way = 10+8=18ways

  4. arindameducationusc
    • one year ago
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    \(\Large\color{red}{Q} \) There are 3 stations A,B,C. 5 routes are going from A to B and 4 routes are going from B to C. Find the number of ways in which a person can go A to C via B. \(\Large\color{Blue}{Ans} \) |dw:1440311727639:dw| Number of ways=5*4=20

  5. arindameducationusc
    • one year ago
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    \(\Large\color{red}{Q} \) A room has 6 doors. In how many ways a man can enter through one door and come out through a different door. \(\Large\color{Blue}{Ans} \) |dw:1440312250191:dw| No. of ways =6*5=30

  6. anonymous
    • one year ago
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    Another good post bro. I'd love to see combination problem too.

  7. anonymous
    • one year ago
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    I remember where there is permutation there is also combination if the question is given a slight twist.

  8. arindameducationusc
    • one year ago
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    \(\large\color{red}{Q} \) Find the no. of four letter words with or without meaning which can be formed with the letters of the word 'Rose' when \(\large\color{red}{1)} \) repetition of the letters is not allowed \(\large\color{red}{2)} \) repetition is allowed \(\large\color{Blue}{Ans1} \) |dw:1440312730296:dw| No. of letters=4*3*2*1=24 This is because when I use one letter, I cannot use that again. In the first box, I can fill any four letters, second box 3(as one already filled in first box), third box 2, and 4th box 1way. \(\large\color{Blue}{Ans2} \) |dw:1440312984349:dw| Each box here can be filled by four ways as repetition allowed. So, Number of letters 4*4*4*4=4^4=256

  9. arindameducationusc
    • one year ago
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    \(\large\color{red}{Q} \) How many no.'s are there b/w 100 & 1000 such that atleast one of their digit is 7? \(\large\color{Blue}{Ans} \) we know there are 10 digits (0,1,2,3,4,5,6,7,8,9) Now, for 3 digit number |dw:1440313369552:dw| here the first box can be filled up in 9 ways, 2nd box in 10 ways and 3rd box in 10 ways. Notice why the first box is filled in 9 ways, cauz if we put zero(0) then it doesn't become a digit number. That's why. Without zero there are 9 digits to fill the first box. so, 9*10*10=900 Now, 3 digit when seven is not present, 0,1,2,3,4,5,6,8,9 (Notice 7 not there) |dw:1440313555444:dw| 8*9*9=648 Now numbers with one digit 7 = 900-648 =252

  10. arindameducationusc
    • one year ago
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    \(\large\color{red}{Q} \) How many no.'s are there b/w 100 and 1000 which have exactly one digit as 7 \(\large\color{Blue}{Ans} \) |dw:1440313804882:dw| In this figure the 7 is fixed in hundredth(BoxA), Tenth(BoxB), Unit(BoxC) position. considering that, we take out the number of ways(in green) Box A=> 1*9*9=81 Box B=> 8*1*9=72 Box C=>8*9*1=72 Total numbers => 81+72+72=225