~~help ASAP assignment 2!!!

**kansairacing** · June 7th 2006, 09:11 PM

can any one do assignment 2 as soon as possible?

when u have answer plz explain how did u get it.....and show every steps plz......thnx very much

thank you u all.....thnx so much

**malaygoel** · June 7th 2006, 09:20 PM

Originally Posted by kansairacing

can any one do assignment 2 as soon as possible?

when u have answer plz explain how did u get it.....and show every steps plz......thnx very much

we have to form six digit numbers taking care of 1,9,0.
for the first digit we can use 7 numbers(2,3,4,5,6,7,8)
for the second digit we can use 8 numbers(0,2,3,4,5,6,7,8)
for the third digit we can use 8 numbers(0,2,3,4,5,6,7,8)
for the fourth digit we can use 8 numbers(0,2,3,4,5,6,7,8)
for the fifth digit we can use 8 numbers(0,2,3,4,5,6,7,8)
for the sixth digit we can use 8 numbers(0,2,3,4,5,6,7,8)
hence total possible numbers = 7*8*8*8*8*8

**CaptainBlack** · June 11th 2006, 09:24 AM

Originally Posted by kansairacing

plz hlep me........i need it ASAP...............

What do you not understand about malaygoel's answer?

RonL

**Soroban** · June 17th 2006, 05:48 AM

Hello, kansairacing!

Is it true that you can't do any of these? . . . sorry to hear it!

Here are a few of them . . .

8. John Finds 10 shirts in his size at a clearance sale.
How many different purchases could he make?

For each of the ten shirts, he has a two-way decision: buy or not-buy.

Therefore, there are: $2^{10} = 1024$ decisions/purchases he can make.

[Note: this includes the decision to buy no shirts.]

10. A school dance has 10 volunteers. .Each dance requires 2 volunteers at the door,
3 volunteers on the floor, and 5 floaters. .If two of the volunteers, John and Tom,
can not work together, in how many ways can the volunteers be assigned?

There are: $\binom{10}{2,3,5}\:=\:\frac{10!}{2!3!5!}\:=\;2520$ ways to make the assignments.

We will count the number of ways that John and Tom are together.
Duct-tape them together; then we have nine "people" to arrange.
. . $\{JT,\;A,\;B,\;C,\;D,\,E,\;F,\;G,\:H\}$

If $JT$ are at the door, the other 8 people can be assigned in
. . $\binom{8}{3,5} \:=\:\frac{8!}{3!5!}\:=\:56$ ways.

If $JT$ are on the floor, the other 8 people can be assigned in
. . $\binom{8}{2,1,5}\:=\:\frac{8!}{2!1!5!}\:=\:126$ ways.

If $JT$ are floaters, the other 8 people can be assigned in
. . $\binom{8}{2,3,3}\:=\:\frac{8!}{2!3!3!}\:=\:560$ ways.

Hence, there are: $56 + 126 + 560\,=\,742$ ways that John and Tom are together.

Therefore, there are: $2520 - 742 \,=\,1778$ ways that they are not together.

*

**Soroban** · June 17th 2006, 10:24 PM

Hello, kansairacing!

Here's #3.
[I assume you know that there is an "E" in MATHEMATICS . . .

Find the number of ways arranging the letters of MATHMATICS if:
(a) there are no restrictions
(b) the first letter must be C
(c) the EVEN-numbered position must remain unchanged.
(d) the letters C and H must be separated

There are 10 letters, including 2 A's, 2 M's, and 2 T's.

$(a)$ With no restrictions, there are: $\frac{10!}{2!2!2!}\;=\;453,600$ ways.

$(b)$ If the first letter must be C, the other nine letters can be arranged in:
. . . $\frac{9!}{2!2!2!}\:=\:45,360$ ways.

$(c)$ If the even-numbered positions are unchanged, we have: $\_\ A\_\ H\_\ A\_\ I\_\ S$
Then the remaining $\{C,M,M,T,T\}$ can have $\frac{5!}{2!2!}\:=\;30$ arrangements.

$(d)$ We will count the way that C and H are together.

Duct-tape them together.
Then we have nine "letters" to arrange: $\{\overline{CH}, A, A, I, M, M, S, T, T\}$
. . These can be arranged in $\frac{9!}{2!2!2!}\:=\:45,360$ ways.

But C and H could have been taped like this: $\{\overline{HC},A,A,I,M,M,S,T,T\}$
. . And these can be arranged in another $45,360$ ways.

Hence, there are: $2 \times 45,360\:=\:90,720$ ways for C and H to be together.

Therefore, there are: $453,600 - 90,720\:=\:362,800$ for C and H to be separated.

**CaptainBlack** · June 17th 2006, 10:36 PM

6. (8 marks) Develop a formula relating $t_{n,r}$ of Pascal's triangle to the terms in row $n-2$ .

For the sake of convenience take:

$ t_{n,r}=0,\ \mbox{for }r>n,\ \mbox{ or } r<1 $

Then the basic relation between the entries in one row and the preceeding
row in Pascal's triangle is:

$ t_{n,r}=t_{n-1,r-1}+t_{n-1,r}\ \ \ \dots (1) $ ,

so applying the rule to the first term of the right hand side of the above:

$ t_{n-1,r-1}=t_{n-2,r-2}+t_{n-2,r-1} $ ,

and to the second term:

$ t_{n-1,r}=t_{n-2,r-1}+t_{n-2,r} $ .

Substituting these into $(1)$ gives:

$ t_{n,r}=t_{n-2,r-2}+2t_{n-2,r-1}+t_{n-2,r} $ .

RonL

**Quick** · June 18th 2006, 04:15 AM

Originally Posted by malaygoel

we have to form six digit numbers taking care of 1,9,0.
for the first digit we can use 7 numbers(2,3,4,5,6,7,8)
for the second digit we can use 8 numbers(0,2,3,4,5,6,7,8)
for the third digit we can use 8 numbers(0,2,3,4,5,6,7,8)
for the fourth digit we can use 8 numbers(0,2,3,4,5,6,7,8)
for the fifth digit we can use 8 numbers(0,2,3,4,5,6,7,8)
for the sixth digit we can use 8 numbers(0,2,3,4,5,6,7,8)
hence total possible numbers = 7*8*8*8*8*8

I disagree, in one of those 6 digits you have to use a zero.

So you would find the number of combinations without zero, which is 7^6=117,649

Then you would subtract that number from the total possible combinations, 7*8^5-7^6=111,727

and that leaves you with the number of six-digit numbers that contain zero

**Soroban** · June 18th 2006, 06:18 AM

Hello again, kansairacing!

Since there is disagreement about #2, here's my solution . . .

2. How many 6-digit numbers are there that include the digit 0
and exclude the digits 1 and 9?

Find the number of six-digit numbers that exclude 1 and 9.
For each of the six digits, there are 8 choices: . $8^6$ numbers.

Among these, how many exclude 0?
For each of the six digits, there are 7 choices: . $7^6$ numbers.

The difference is number of 6-digit numbers that exclude 1 and 9 and include 0:

. . . $8^6 - 7^6\:=\:262,144 - 117,649\:=\:144,495$

**Quick** · June 18th 2006, 07:50 AM

Originally Posted by Soroban

Hello again, kansairacing!

Since there is disagreement about #2, here's my solution . . .

Find the number of six-digit numbers that exclude 1 and 9.
For each of the six digits, there are 8 choices: . $8^6$ numbers.

Among these, how many exclude 0?
For each of the six digits, there are 7 choices: . $7^6$ numbers.

The difference is number of 6-digit numbers that exclude 1 and 9 and include 0:

. . . $8^6 - 7^6\:=\:262,144 - 117,649\:=\:144,495$

The first digit can't have 0, which leaves only seven different numbers possible for the first digit, because no number greater than 1 starts with 0.

**kansairacing** · June 18th 2006, 07:34 PM

how do u do nubmer 7, 9?

i dont understand the question...

**Soroban** · June 19th 2006, 02:42 AM

Hello, Quick!

The first digit can't have 0 . . .

You're right . . . I forgot that convention . . . *blush*
I'll try again.

2. How many 6-digit numbers are there that include the digit 0
and exclude the digits 1 and 9?

Find the number of six-digit numbers that exclude 1 and 9.

The first digit cannot be 0; there are 7 choices.
Each of the remaining five digits has 8 choices.
. . Hence, there are: $7\cdot8^5$ numbers.

Among these, how many exclude 0?
. . For each of the six digits, there are 7 choices: . $7^6$ numbers.

The difference is number of 6-digit numbers that exclude 1 and 9 and include 0:

. . . $7\cdot8^5 - 7^6\:=\:229,376 - 117,649\:=\:111,727$

**Soroban** · June 19th 2006, 03:47 AM

Hello, kansairacing!

It would be gratifying to get some feedback . . .
. . Are we helping at all?
. . Is there any learning taking place?
. . Or are we just doing your homework for you?

7) Suppose you are designing a remote control that uses short or long pulses
of infrared light to send control signals to a device.
(a) How many different codes can you defined using two or three pulses?
(b) Explain how the Rule of Product and the Rule of Sum apply in your calculation.

(a) The very least you could do is list the possibilities . . . right?

With two pulses: $LL,\;LS,\;SL,\;SS$ . . . 4 signals.
With three pulses: $LLL,\;LLS,\,LSL,\,LSS,\;SLL,\;SLS,\;SSL,\;SSS$ . . . 8 signals.
. . Answer: $4 + 8 \,=\,12$ signals.

(b) Two pulses: for each of the two pulses, there are two choices, L or S.
. . Hence, there are: $2^2 = 4$ two-pulse signals.
Three pulses: for each of the three pulses, there are two choices.
. . Hence, there are: $2^3 = 8$ three-pulse signals.
These two steps use the Rule of Product.

Therefore, for two or three pulses, there are: $4 + 8 \,=\,12$ signals.
This step uses the Rule of Sum.

9) Suppose the Canadian Embassy in the Netherlands has 28 employees.
Eight of the employees speak German and ten speak Dutch.
If two empolyees speak both German and Dutch,
how many of the employees speak neither German nor Dutch?
Illustrate your answer with a Venn diagram.

Code:

      * - - - - - - - - - - - - - - *
      |                             |
      |     * - - - - - *           |
      |     | G   6     |           |
      |     |     * - - + - - *     |
      |     |     |  2  |     |     |
      |     * - - + - - *     |     |
      |           |     8   D |     |
      |           * - - - - - *     |
      |   12                        |
      * - - - - - - - - - - - - - - *

The large rectangle represents the 28 employees.

We are told that 2 speak German and Dutch.
The "2" goes in the region common to the German square and the Dutch square.

Since 8 of them speak German, there are 6 who speak only German.
Since 10 of them speak Dutch, there are 8 who speak only Dutch.

We have accounted for: $6 + 2 + 8 \,= \,16$ of the employees.

Therefore, the other $28 - 16 \,=\,12$ are those who speak neither German nor Dutch.

**CaptainBlack** · June 19th 2006, 05:09 AM

Originally Posted by Soroban

Hello, kansairacing!

It would be gratifying to get some feedback . . .
. . Are we helping at all?
. . Is there any learning taking place?

There certainly is on the part of the helpers

. . Or are we just doing your homework for you?

**kansairacing** · June 19th 2006, 04:51 PM

THank you all...............i love u all very much........thnx

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