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Thread: Permutations Questions

  1. #1
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    Exclamation Permutations Questions

    1. I have 3 Maths books, 2 Science books and 4 English books which are all different. how many ways can I arrange the books on a shelf so that books of the same subject are put together?

    2. How many different 5 digit numbers that are odd and less than 50 000 can be formed using the digits 2,3,4,5,6,7,8 without repetition?

    3. Find the number of different 7-digit numbers (not beginning with zero) that can be formed using the digits 0-9 without repetition if the number is a multiple of 5.
    Ans:
    1. 1728
    2. 480
    3. 329280

    Any help please? I need to turn this is in by tomorrow. TIA.
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  2. #2
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    1. First, treat the books that must stay together- the books of one subject- as a single object. There are three such objects- "Math books", "Science books", "English books" so there are 3!= 6 ways to order the subjects. Since there are 3 Math books, there are 3!= 6 ways to order them within their group. Since there are there are 2!= 2 ways to order them. Since there are 4 English books, there are 4!= 24 ways to order them.

    Putting that together, there are (6)(6)(2)(24)= 1728 ways to order all 9 books. That is considerably smaller than 7!= 5040 ways of ordering all 7 books without any restrictions.

    2. Because the number must be odd, the "ones" digit must be odd: 3, 5, or 7.
    Because the number must be less than 5000, the "thousands" digit must be one of 2, 3, or 4.
    We are going to need to break this into "cases":

    If the ones digit is 3, the thousands digit must be either 2 or 4. That leaves 5 digits from which to choose the remaining, middle, 3 digits: 5!/(5-3)!= 5!/2!= 60. That gives 2(60)= 120 possibilities for this case. (That "2" is for the two choices for thousands digit.)

    If the ones digit is either 5 or 7, the thousands digit can be one of 2, 3, or 4. That leaves 5 digits from which to choose the remaining, middle digits. There are, as before, 60 ways to do that. That gives 2(3)(60)= 360 possibilities for this case. Putting those together, there are 120+ 360= 480 ways to do this.

    3. A number is divisible by 5 if and only if its "ones" digit is 0 or 5. Taking either one of those leaves 9 digits for the first 6 digits. The leading digit cannot be 0 so that leaves 8 digits for the leading digit. That, finally, leaves 8 digits for the remaining, middle 5 digits: there are 8!/(8-5)!= 8!/2!= 20160 ways to do that. There are, then (2)(8)(20160)= 322560 ways to do this.
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  3. #3
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    Hello, smmxwell!

    I don't agree with two of their answers.


    1. I have 3 Maths books, 2 Science books and 4 English books which are all different.
    How many ways can I arrange the books on a shelf so that books of the same subject are together?

    The 3 Math books can be placed together in: .$\displaystyle 3! = 6$ ways.
    The 2 Science books can be placed together in: .$\displaystyle 2! = 2$ ways.
    The 4 English books can be placed together in: .$\displaystyle 4! = 24$ ways.

    The three groups of books can be placed together in: .$\displaystyle 3! = 6$ ways.

    Therefore, there are: .$\displaystyle 6 \times 2 \times 24 \times 6 \:=\:\boxed{1728}$ ways.




    2. How many different 5 digit numbers that are odd and less than 50,000
    can be formed using the digits 2,3,4,5,6,7,8 without repetition?
    We have 5-digit numbers: the first digit is 2, 3, or 4 ... the last digit is 3, 5, or 7.


    We must consider two possible cases:

    (1) The last digit is 3: .$\displaystyle \_\;\_\;\_\;\_\;3$
    . . .The first digit can be 2 or 4: .$\displaystyle 2$ choices.
    . . .The other 3 digit are chosen from the remaining 5 digit: .$\displaystyle _5P_3 \:=\: 120$ choices.
    . . Hence, there are: .$\displaystyle 2 \times 120 \:=\:240$ ways

    (2) The last digit is 5 or 7: .$\displaystyle 2$ choices.
    . . .The first digit can be 2, 3 or 4: .$\displaystyle 3$ choices.
    . . .The other 3 digits are chosen from the remaining 5 digits: .$\displaystyle _5P_3 = 120$ choices.
    . . .Hence, there re: .$\displaystyle 2\times 3\times 120 \:=\:720$ ways.

    Therefore, there are: .$\displaystyle 240 + 720 \:=\:\boxed{960}$ odd numbers less than 50,000.




    3. Find the number of different 7-digit numbers (not beginning with zero)
    that can be formed using the digits 0-9 without repetition if the number is a multiple of 5.
    We have 7-digit numbers ending in 0 or 5.


    We must consider two possible cases:

    (1) The last digit is 0: .$\displaystyle \_\;\_\;\_\;\_\:\_\;\_\;0$
    . . .Then the other 6 digits are chosen from the remaining 9 digits: .$\displaystyle _9P_6 \:=\:60,\!480$ ways.
    . . .There are: .$\displaystyle 60,\!480$ seven-digit numbers ending in 0.

    (2) The last digit is 5: .$\displaystyle \_\;\_\;\_\;\_\;\_\;\_\;5$
    . . .The first digit must not be zero: .$\displaystyle 8$ choices.
    . . .Then the other 5 digits are chosen from the remaining 8 digits: .$\displaystyle _8P_5\:=\:6,\!720$ ways.
    . . .There are: .$\displaystyle 8 \times 6,\!720 \:=\:53,\!760$ seven-digit numbers ending in 5.

    Therefore, there are: .$\displaystyle 60,\!480 + 53,\!760 \:=\:\boxed{114,\!240}$ seven-digit multiples of 5.

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  4. #4
    Senior Member pankaj's Avatar
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    1)The books of different subjects can be arranged among themselves in $\displaystyle (3!)(2!)(4!)$ ways and the three subjects can arrange themselves in $\displaystyle (3!)$ways.
    Therefore total number of ways=$\displaystyle (3!)(2!)(4!)(3!)=1728$

    2)Keeping 2 or 4 in the first place
    (2,4)_ _ _ _

    Last place must contain an odd digit(3 choices) and the first place must be either 2 or 4 (hence 2 ways).The remaining places can be filled in 5,4 and 3 ways.
    The number of choices are (2)(3)(4)(5)(3)=360

    Now fixing 3 in the first place
    3_ _ _ _

    Last place has only 2 choices since the first place is occupied by an odd digit.The remaining places can be filled 5,4 and 3 ways.
    The number of choices=(1)(3)(4)(5)(2)=120

    Therefore total number of numbers=360+120=480

    3)Fix 0 in the unit digit's place.The rest of the digits can be arranged in 9.8.7.6.5.4 ways and when you fix 5 in the unit digit's place the rest of the digits can be arranged in 8.8.7.6.5.4.
    Therefore total number of required numbers=9.8.7.6.5.4+8.8.7.6.5.4
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