At a high school assembly, the principal plans to give out 3 awards for extraordinary achievement. The nominees for the awards are the 3 most highly rated teachers and the 5 students with the most improved GPAs. How many different groups of award recipients can be chosen in which at least one recipient is a teacher? (Two groups of award recipients are considered different if at least one group member is different.)
(A) 20
(B) 30
(C) 46
(D) 60
(E) 120
OA is C
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are you sure the question is not missing information ? How many student and teachers are there ?
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whenever you see at least in these types of questions do the following (if my answer is correct):
1- total possible 3 groups of 3 out of 8 -----> 8!/3!(8-3)! = 56
2- total number of groups of 3 without any teacher-------> 5!/3!(5-3)! = 10
S0, 56-10 = 46
1- total possible 3 groups of 3 out of 8 -----> 8!/3!(8-3)! = 56
2- total number of groups of 3 without any teacher-------> 5!/3!(5-3)! = 10
S0, 56-10 = 46
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Cool man .. you're doing well now.. tell me what else do you know in Comb., Perm. ,and Prob?heshamelaziry wrote:whenever you see at least in these types of questions do the following (if my answer is correct):
1- total possible 3 groups of 3 out of 8 -----> 8!/3!(8-3)! = 56
2- total number of groups of 3 without any teacher-------> 5!/3!(5-3)! = 10
S0, 56-10 = 46
Abdulla
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With these 4 problems that show 3 concepts, this is all I know about per, prob, comb
Q) John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter ?
1/8
1/6
2/9
5/18
1/3
ANSWER: Total Players to choose from : 9
No of players to choose : 5
Restriction : Team with John and Peter
- - - - -
The order in which the team members are selected doesn't matter.
1) No of ways of selecting John : 1
2) No of ways of selecting Peter : 1
3) No of ways of selecting the remaining 7i / 3i( 7-3i) = 35
4) 1 *2*3 = TOTAL DESIRED OUTCOME = 35
5) TOTAL POSSIBLE OUTCOMES = 9i / 5i( 9-5i ) = 126
SOLUTION = 4 / 5 = 5/18
Q) 9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
The guards can be chosen in 6C3 ways.
The forwards can be chosen in 3C2 ways.
6C3 * 3C2
6C3 = 20
3C2 = 3
20 * 3 = 60
Q) "If you flip a coin five times, what's the probability you get exactly three heads ?
The answer can be found in ten seconds- it's 5C3/2^5 = 5/16
Q) A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?
Answer:
We can treat this exactly as we would a coin flip question; think of it as:
"If a fair coin is flipped 4 times, what's the probability of getting exactly 2 heads?"
There's an easy to use formula for coin flip (and pseudo-coin flip) questions:
Probability of getting exactly k results out of n flips = nCk/2^n
(nCk = n!/k!(n-k)!, the combinations formula.)
In this question, we have 4 flips and we want 2 heads, so n=4 and k=2:
4C2/2^4 = (4!/2!2!)/16 = (24/4)/16 = 6/16 = 3/8
Q) John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter ?
1/8
1/6
2/9
5/18
1/3
ANSWER: Total Players to choose from : 9
No of players to choose : 5
Restriction : Team with John and Peter
- - - - -
The order in which the team members are selected doesn't matter.
1) No of ways of selecting John : 1
2) No of ways of selecting Peter : 1
3) No of ways of selecting the remaining 7i / 3i( 7-3i) = 35
4) 1 *2*3 = TOTAL DESIRED OUTCOME = 35
5) TOTAL POSSIBLE OUTCOMES = 9i / 5i( 9-5i ) = 126
SOLUTION = 4 / 5 = 5/18
Q) 9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
The guards can be chosen in 6C3 ways.
The forwards can be chosen in 3C2 ways.
6C3 * 3C2
6C3 = 20
3C2 = 3
20 * 3 = 60
Q) "If you flip a coin five times, what's the probability you get exactly three heads ?
The answer can be found in ten seconds- it's 5C3/2^5 = 5/16
Q) A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?
Answer:
We can treat this exactly as we would a coin flip question; think of it as:
"If a fair coin is flipped 4 times, what's the probability of getting exactly 2 heads?"
There's an easy to use formula for coin flip (and pseudo-coin flip) questions:
Probability of getting exactly k results out of n flips = nCk/2^n
(nCk = n!/k!(n-k)!, the combinations formula.)
In this question, we have 4 flips and we want 2 heads, so n=4 and k=2:
4C2/2^4 = (4!/2!2!)/16 = (24/4)/16 = 6/16 = 3/8
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Thanks hesham it was very helpful ..heshamelaziry wrote:With these 4 problems that show 3 concepts, this is all I know about per, prob, comb
Q) John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter ?
1/8
1/6
2/9
5/18
1/3
ANSWER: Total Players to choose from : 9
No of players to choose : 5
Restriction : Team with John and Peter
- - - - -
The order in which the team members are selected doesn't matter.
1) No of ways of selecting John : 1
2) No of ways of selecting Peter : 1
3) No of ways of selecting the remaining 7i / 3i( 7-3i) = 35
4) 1 *2*3 = TOTAL DESIRED OUTCOME = 35
5) TOTAL POSSIBLE OUTCOMES = 9i / 5i( 9-5i ) = 126
SOLUTION = 4 / 5 = 5/18
Q) 9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
The guards can be chosen in 6C3 ways.
The forwards can be chosen in 3C2 ways.
6C3 * 3C2
6C3 = 20
3C2 = 3
20 * 3 = 60
Q) "If you flip a coin five times, what's the probability you get exactly three heads ?
The answer can be found in ten seconds- it's 5C3/2^5 = 5/16
Q) A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?
Answer:
We can treat this exactly as we would a coin flip question; think of it as:
"If a fair coin is flipped 4 times, what's the probability of getting exactly 2 heads?"
There's an easy to use formula for coin flip (and pseudo-coin flip) questions:
Probability of getting exactly k results out of n flips = nCk/2^n
(nCk = n!/k!(n-k)!, the combinations formula.)
In this question, we have 4 flips and we want 2 heads, so n=4 and k=2:
4C2/2^4 = (4!/2!2!)/16 = (24/4)/16 = 6/16 = 3/8
Abdulla
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Could you help with this:
Q. What is the units digit of a^36?
a) a^2 has 9 as the units digit
b) a^3 has 3 has the units digit
Q. What is the units digit of a^36?
a) a^2 has 9 as the units digit
b) a^3 has 3 has the units digit
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I don't know 2010gmat. I am poor in this subject. I just memorize question patterns, unfortunately. I never took a class that covers all this stuff.2010gmat wrote:great ques hesham...
if the question says at least 3 heads in 5 tosses...
then??
will it be
5c0 + 5c1 + 5c2 + 5c3)/2^5 ???i.e 30/32
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Sure, I think the answer is Dheshamelaziry wrote:Could you help with this:
Q. What is the units digit of a^36?
a) a^2 has 9 as the units digit
b) a^3 has 3 has the units digit
All what you need to do is to find the pattern. from the first statement we know that a^2 = 9 , so try to continue to get the pattern ..
If the unit digit of a^2 = 9 then the unit digit of a^3 = 1 ( 9 * 9 = 81) and so on.. you will notice that the unit digit will be either 9 or 1 ... so we can get the answer.. Sufficient.
Statement 2 ... same steps .. Sufficient.
Abdulla
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Abdulla wrote:Sure, I think the answer is Dheshamelaziry wrote:Could you help with this:
Q. What is the units digit of a^36?
a) a^2 has 9 as the units digit
b) a^3 has 3 has the units digit
All what you need to do is to find the pattern. from the first statement we know that a^2 = 9 , so try to continue to get the pattern ..
If the unit digit of a^2 = 9 then the unit digit of a^3 = 1 ( 9 * 9 = 81) and so on.. you will notice that the unit digit will be either 9 or 1 ... so we can get the answer.. Sufficient.
Statement 2 ... same steps .. Sufficient.
Apologize for the bother. I know how to do this if i have a number, but what is the concept behind a^3 = 81 ? a^2 = 9 means that a =3 so why not a^3 = 27 ?
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Sorry I missed it up, but still the concept is the same and the correct answer is the same.heshamelaziry wrote:Abdulla wrote:Sure, I think the answer is Dheshamelaziry wrote:Could you help with this:
Q. What is the units digit of a^36?
a) a^2 has 9 as the units digit
b) a^3 has 3 has the units digit
All what you need to do is to find the pattern. from the first statement we know that a^2 = 9 , so try to continue to get the pattern ..
If the unit digit of a^2 = 9 then the unit digit of a^3 = 1 ( 9 * 9 = 81) and so on.. you will notice that the unit digit will be either 9 or 1 ... so we can get the answer.. Sufficient.
Statement 2 ... same steps .. Sufficient.
Apologize for the bother. I know how to do this if i have a number, but what is the concept behind a^3 = 81 ? a^2 = 9 means that a =3 so why not a^3 = 27 ?
Just try to get the pattern .. the key is that the unit digit will keep repeated .. for example..
3^1=3 3^2= 9 3^3=27 3^4 = 81 3^5= 243 ... so If we continue we definitely will find the unit digit for a^36... sorry for my bad explanations..
Abdulla
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Hi Hesham,
Im gonna help you with the sum to find a^36. Now see we are only concerned with the units digit here.. so lets see which square(s) gives us the units digit 9, n which cube(s) give us 3 in the units digit. So here go the squares - 2 = 4, 3 = 9, 4 = 16, 5 = 25, 6 = 36, 7 = 49.thus only 3 and 7 end in units digit 9, when they are squared. Therefore, lets see the cubes of 3 and 7 to decide which no. it is. 3^3 = 27, so this does not work coz we need 3 in the units place. Now take 7, 7^3 = 49 x 7 = 343. So bingo.. 7 is our no.
Now the next step is to find a pattern.... for 7^n... 7^1 = 7, 7^2 = 49, 7^3 = 343, 7 ^4 = here the units digit ends in 1 as 7 x 3 = 21, next multiply the units digit by 7..so 7^5 = units digit is 7. Thus every 4th one ends in 1, and 36 being a multiple of 4 also ends in 1, or has 1 in the units place.
Im gonna help you with the sum to find a^36. Now see we are only concerned with the units digit here.. so lets see which square(s) gives us the units digit 9, n which cube(s) give us 3 in the units digit. So here go the squares - 2 = 4, 3 = 9, 4 = 16, 5 = 25, 6 = 36, 7 = 49.thus only 3 and 7 end in units digit 9, when they are squared. Therefore, lets see the cubes of 3 and 7 to decide which no. it is. 3^3 = 27, so this does not work coz we need 3 in the units place. Now take 7, 7^3 = 49 x 7 = 343. So bingo.. 7 is our no.
Now the next step is to find a pattern.... for 7^n... 7^1 = 7, 7^2 = 49, 7^3 = 343, 7 ^4 = here the units digit ends in 1 as 7 x 3 = 21, next multiply the units digit by 7..so 7^5 = units digit is 7. Thus every 4th one ends in 1, and 36 being a multiple of 4 also ends in 1, or has 1 in the units place.
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Is the answer b or d in your opinion ?Pooja Bhula wrote:Hi Hesham,
Im gonna help you with the sum to find a^36. Now see we are only concerned with the units digit here.. so lets see which square(s) gives us the units digit 9, n which cube(s) give us 3 in the units digit. So here go the squares - 2 = 4, 3 = 9, 4 = 16, 5 = 25, 6 = 36, 7 = 49.thus only 3 and 7 end in units digit 9, when they are squared. Therefore, lets see the cubes of 3 and 7 to decide which no. it is. 3^3 = 27, so this does not work coz we need 3 in the units place. Now take 7, 7^3 = 49 x 7 = 343. So bingo.. 7 is our no.
Now the next step is to find a pattern.... for 7^n... 7^1 = 7, 7^2 = 49, 7^3 = 343, 7 ^4 = here the units digit ends in 1 as 7 x 3 = 21, next multiply the units digit by 7..so 7^5 = units digit is 7. Thus every 4th one ends in 1, and 36 being a multiple of 4 also ends in 1, or has 1 in the units place.