1. ## 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.

2. 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.

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

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

Therefore, there are: . $6 \times 2 \times 24 \times 6 \:=\:\boxed{1728}$ 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: . $\_\;\_\;\_\;\_\;3$
. . .The first digit can be 2 or 4: . $2$ choices.
. . .The other 3 digit are chosen from the remaining 5 digit: . $_5P_3 \:=\: 120$ choices.
. . Hence, there are: . $2 \times 120 \:=\:240$ ways

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

Therefore, there are: . $240 + 720 \:=\:\boxed{960}$ 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: . $\_\;\_\;\_\;\_\:\_\;\_\;0$
. . .Then the other 6 digits are chosen from the remaining 9 digits: . $_9P_6 \:=\:60,\!480$ ways.
. . .There are: . $60,\!480$ seven-digit numbers ending in 0.

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

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

4. 1)The books of different subjects can be arranged among themselves in $(3!)(2!)(4!)$ ways and the three subjects can arrange themselves in $(3!)$ways.
Therefore total number of ways= $(3!)(2!)(4!)(3!)=1728$

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