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 April 8th, 2011, 12:24 AM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Probability question(hard) 1.Find the number of ways in which a committee of 4 can be chosen from 6 boys and 6 girls a)if either the oldest boy or oldest girls must be included but not the both (answer:240) 2.Find in factor form the number of ways in which 20 boys can be arranged in a line from right to left so that no two of three particular boys will be standing next to each other. 3.A certain test consists of seven questions,to each of which a candidate must give one of three possible answers.According to the answer that he chooses,the candidate must score 1,2 or 3 marks for each of the seven questions.In how many different ways can a candidate score exactly 18 marks in the test? (answer:77) Please help me to solve these question step by step and explain to me.i am confuse about these question and stuck in it.i need help.
 April 8th, 2011, 12:49 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Probability question(hard) 1. (i) Suppose the eldest boy is included. You can cross out the eldest girl and then pick the other three members. (ii) Suppose the eldest girl is included. You can cross out the eldest boy and then pick the other three members. Please now show your attempt to complete each part.
April 8th, 2011, 06:50 AM   #3
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Re: Probability question(hard)

Hello, hoyy1kolko!

Quote:
 2. Find in factor form the number of ways in which 20 boys can be arranged in a line from right to left so that no two of three particular boys will be standing next to each other.

$\text{Suppose the three particular boys are }A,\,B,\,C\text{ and there are 17 Others.}$

$\text{Place the 17 Others in a row. \;There are: }17!\text{ possible orders.}$

$\text{Insert a space before, after and between them.}$
[color=beige]. . [/color]$\_\;x\;\_\;x\;\_\;x\,\_\;.\;.\;.\;\_\;x\;\_$

$\text{Choose 3 of the 18 spaces to place the 3 particular boys.}$

[color=beige]. . [/color]$\text{There are: }\:\begin{Bmatrix}18\text{ choices for }A \\ \\ \\17\text{ choices for }B\\ \\ \\ 16\text{ choices for }C \end{Bmatrix}$

$\text{Hence, there are: }\:18\cdot17\cdot16\text{ ways to place }A,\,B,\text{ and }C.$

$\text{Therefore, there are: }\17!)(18\cdot17\cdot16)\text{ possible arrangements.}" />

 April 8th, 2011, 07:33 AM #4 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability question(hard) my solution for question 1 i) Suppose the eldest boy is included. You can cross out the eldest girl and then pick the other three members. 6C1 x 6C3=120 (ii) Suppose the eldest girl is included. You can cross out the eldest boy and then pick the other three members. 6C1 x 6C3=120 then the answer 120+120=240 . is it the right way to get the answer.?
April 8th, 2011, 01:42 PM   #5
Math Team

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From: Lexington, MA

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Re: Probability question(hard)

Hello again, hoyy1kolko!

Quote:
 3. A certain test consists of seven questions, to each of which a candidate must give one of three possible answers. According to the answer that he chooses, the candidate must score 1,2 or 3 marks for each of the seven questions. In how many different ways can a candidate score exactly 18 marks in the test? [color=beige] .[/color][Answer: 77]

$\text{Let: }\;\begin{Bmatrix} a=&\text{number of 1-point questions} \\ b=&\text{number of 2-point questions} \\ c=&\text{number of 3-point questions} \end{Bmarix} \;\;\text{where: }\;\begin{Bmatrix}a\,+\,b\,+\,c=&7 \\ a\,+\,2b\,+\,3c=&18 \end{Bmatrix}=$

$\text{We find that there are only two solutions: }\;(a,\,b,\,c) \;=\;(1,\,1,\,5)\,\text{ and }\,(0,\,3,\,4)$

$\text{The selection }\,(1,\,1,\,5)\,\text{ can be made in }\,\frac{7!}{\,1!\,1!\,5!} \:=\: 42\text{ ways.}$

$\text{The selection }\,(0,\,3,\,4)\,\text{ can be made in }\,\frac{7!}{0!\,3!\,4!} \:=\:35\text{ ways.}$

$\text{Therefore, there are: }\:42\,+\,35\:=\:77\text{ ways to score exactly 18 points.}$

April 8th, 2011, 06:12 PM   #6
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Re: Probability question(hard)

Quote:
 Originally Posted by hoyy1kolko my solution for question 1 i) Suppose the eldest boy is included. You can cross out the eldest girl and then pick the other three members. 6C1 x 6C3=120 (ii) Suppose the eldest girl is included. You can cross out the eldest boy and then pick the other three members. 6C1 x 6C3=120 then the answer 120+120=240 . is it the right way to get the answer.?
You have:
ELDEST BOY, ELDEST GIRL
BOY BOY BOY BOY BOY
GIRL GIRL GIRL GIRL GIRL

If you have chosen the eldest boy, you cross out the eldest girl then choose 3 out of the 10 remaining.
10 choose 3 = 10! / 7!3! = 120
If you chose the eldest girl, you cross out the eldest boy, and choose 3 out of the 10 remaining.
Again, 120

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