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

http://i74.photobucket.com/albums/i2...4/scan0002.jpg

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

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- Jun 7th 2006, 09:11 PMkansairacing~~help ASAP assignment 2!!!
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

http://i74.photobucket.com/albums/i2...4/scan0002.jpg

thank you u all.....thnx so much - Jun 7th 2006, 09:20 PMmalaygoelQuestion no.2Quote:

Originally Posted by**kansairacing**

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 - Jun 11th 2006, 09:24 AMCaptainBlackQuote:

Originally Posted by**kansairacing**

RonL - Jun 17th 2006, 05:48 AMSoroban
Hello, kansairacing!

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

Here are a few of them . . .

Quote:

8. John Finds 10 shirts in his size at a clearance sale.

How many different purchases could he make?

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

[Note: this includes the decision to buy__no__shirts.]

Quote:

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?

We will count the number of ways that John and Tom**are**together.

Duct-tape them together; then we have nine "people" to arrange.

. . $\displaystyle \{JT,\;A,\;B,\;C,\;D,\,E,\;F,\;G,\:H\}$

If $\displaystyle JT$ are at the door, the other 8 people can be assigned in

. . $\displaystyle \binom{8}{3,5} \:=\:\frac{8!}{3!5!}\:=\:56$ ways.

If $\displaystyle JT$ are on the floor, the other 8 people can be assigned in

. . $\displaystyle \binom{8}{2,1,5}\:=\:\frac{8!}{2!1!5!}\:=\:126$ ways.

If $\displaystyle JT$ are floaters, the other 8 people can be assigned in

. . $\displaystyle \binom{8}{2,3,3}\:=\:\frac{8!}{2!3!3!}\:=\:560$ ways.

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

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

* - Jun 17th 2006, 10:24 PMSoroban
Hello, kansairacing!

Here's #3.

[I assume you know that there is an "E" in MATH**E**MATICS . . .

Quote:

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

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

$\displaystyle (b)$ If the first letter must be C, the other nine letters can be arranged in:

. . . $\displaystyle \frac{9!}{2!2!2!}\:=\:45,360$ ways.

$\displaystyle (c)$ If the even-numbered positions are unchanged, we have: $\displaystyle \_\ A\_\ H\_\ A\_\ I\_\ S$

Then the remaining $\displaystyle \{C,M,M,T,T\}$ can have $\displaystyle \frac{5!}{2!2!}\:=\;30$ arrangements.

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

Duct-tape them together.

Then we have nine "letters" to arrange: $\displaystyle \{\overline{CH}, A, A, I, M, M, S, T, T\}$

. . These can be arranged in $\displaystyle \frac{9!}{2!2!2!}\:=\:45,360$ ways.

But C and H could have been taped like this: $\displaystyle \{\overline{HC},A,A,I,M,M,S,T,T\}$

. . And these can be arranged in*another*$\displaystyle 45,360$ ways.

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

Therefore, there are: $\displaystyle 453,600 - 90,720\:=\:362,800$ for C and H to be separated. - Jun 17th 2006, 10:36 PMCaptainBlack

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

For the sake of convenience take:

$\displaystyle

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:

$\displaystyle

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:

$\displaystyle

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

$,

and to the second term:

$\displaystyle

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

$.

Substituting these into $\displaystyle (1)$ gives:

$\displaystyle

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

$.

RonL - Jun 18th 2006, 04:15 AMQuickQuote:

Originally Posted by**malaygoel**

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 - Jun 18th 2006, 06:18 AMSoroban
Hello again, kansairacing!

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

Quote:

2. How many 6-digit numbers are there that include the digit 0

and exclude the digits 1 and 9?

For each of the six digits, there are 8 choices: .$\displaystyle 8^6$ numbers.

Among these, how many*exclude*0?

For each of the six digits, there are 7 choices: .$\displaystyle 7^6$ numbers.

The difference is number of 6-digit numbers that exclude 1 and 9 and__include__0:

. . . $\displaystyle 8^6 - 7^6\:=\:262,144 - 117,649\:=\:144,495$ - Jun 18th 2006, 07:50 AMQuickQuote:

Originally Posted by**Soroban**

- Jun 18th 2006, 07:34 PMkansairacing
how do u do nubmer 7, 9?

i dont understand the question... - Jun 19th 2006, 02:42 AMSoroban
Hello, Quick!

Quote:

The first digit can't have 0 . . .

I'll try again.

Quote:

2. How many 6-digit numbers are there that include the digit 0

and exclude the digits 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: $\displaystyle 7\cdot8^5$ numbers.

Among these, how many*exclude*0?

. . For each of the six digits, there are 7 choices: .$\displaystyle 7^6$ numbers.

The difference is number of 6-digit numbers that exclude 1 and 9 and__include__0:

. . . $\displaystyle 7\cdot8^5 - 7^6\:=\:229,376 - 117,649\:=\:111,727$ - Jun 19th 2006, 03:47 AMSoroban
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?

Quote:

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.

*least*you could do is**list**the possibilities . . . right?

With two pulses: $\displaystyle LL,\;LS,\;SL,\;SS$ . . . 4 signals.

With three pulses: $\displaystyle LLL,\;LLS,\,LSL,\,LSS,\;SLL,\;SLS,\;SSL,\;SSS$ . . . 8 signals.

. . Answer: $\displaystyle 4 + 8 \,=\,12$ signals.

(b) Two pulses: for each of the two pulses, there are two choices, L or S.

. . Hence, there are: $\displaystyle 2^2 = 4$ two-pulse signals.

Three pulses: for each of the three pulses, there are two choices.

. . Hence, there are: $\displaystyle 2^3 = 8$ three-pulse signals.

These two steps use the Rule of Product.

Therefore, for two**or**three pulses, there are: $\displaystyle 4 + 8 \,=\,12$ signals.

This step uses the Rule of Sum.

Quote:

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 |

* - - - - - - - - - - - - - - *

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: $\displaystyle 6 + 2 + 8 \,= \,16$ of the employees.

Therefore, the other $\displaystyle 28 - 16 \,=\,12$ are those who speak neither German nor Dutch. - Jun 19th 2006, 05:09 AMCaptainBlackQuote:

Originally Posted by**Soroban**

Quote:

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

- Jun 19th 2006, 04:51 PMkansairacing
THank you all...............i love u all very much........thnx