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How many ways are there for a bank to choose n students?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How many ways to distribute 6 notepads, 7 pencils and 8 markers?How many ways are there to do this so that no officer picks $3$ students from the same high school?A bank has to give 5 positions for 15 candidatesCombinatorics: How many ways are there to distribute zero to thirteen distinct cards to four distinct players?How many ways are there to choose 10 coins with at least 3 nickels but no more than 2 quarters?How many ways are there to distribute three different pens and nineteen identical pencils…?How many ways can $26$ students be distributed.Short Combinatorics problem.How many ways are there to choose from a deck of cards?How many different ways can a group of students be hired to work a survey?
$begingroup$
A bank comes to campus and interviews each person one at a time. After each interview they decide to hire that person or not.
i. There are n students. How many decisions does the bank make?
ii. How many ways are there to choose students to get jobs at the bank?
I am confused about this. I think on the first one that maybe we let k be the number of decisions the bank makes. Since there are n students and the bank makes k decisions, we have $binomnk$. And isn't the second one identical to the first one? Or maybe it's just $2^n$ ways to choose students to offer jobs at the bank?
combinatorics discrete-mathematics
asked Mar 27 at 4:59
usukidollusukidoll
1,1761033
$endgroup$
add a comment |
$begingroup$
A bank comes to campus and interviews each person one at a time. After each interview they decide to hire that person or not.
i. There are n students. How many decisions does the bank make?
ii. How many ways are there to choose students to get jobs at the bank?
I am confused about this. I think on the first one that maybe we let k be the number of decisions the bank makes. Since there are n students and the bank makes k decisions, we have $binomnk$. And isn't the second one identical to the first one? Or maybe it's just $2^n$ ways to choose students to offer jobs at the bank?
combinatorics discrete-mathematics
asked Mar 27 at 4:59
usukidollusukidoll
1,1761033
$endgroup$
3
$begingroup$
You think too complicatedly. They make a decision for every student they interview. They do $n$ interviews. So they make $n$ decisions.
$endgroup$
– amsmath
Mar 27 at 5:06
$begingroup$
I just realized what happened. For i, the bank can either accept or reject a student after the interview so they have $2^n$ decisions for ii, the student can decision whether or not to accept or decline the bank's job offer, so it's also $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:22
$begingroup$
typos... I meant for ii, the student can decide whether or not to accept or decline the bank's job offer so it's $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:34
add a comment |
$begingroup$
A bank comes to campus and interviews each person one at a time. After each interview they decide to hire that person or not.
i. There are n students. How many decisions does the bank make?
ii. How many ways are there to choose students to get jobs at the bank?
I am confused about this. I think on the first one that maybe we let k be the number of decisions the bank makes. Since there are n students and the bank makes k decisions, we have $binomnk$. And isn't the second one identical to the first one? Or maybe it's just $2^n$ ways to choose students to offer jobs at the bank?
combinatorics discrete-mathematics
asked Mar 27 at 4:59
usukidollusukidoll
1,1761033
$endgroup$
A bank comes to campus and interviews each person one at a time. After each interview they decide to hire that person or not.
i. There are n students. How many decisions does the bank make?
ii. How many ways are there to choose students to get jobs at the bank?
I am confused about this. I think on the first one that maybe we let k be the number of decisions the bank makes. Since there are n students and the bank makes k decisions, we have $binomnk$. And isn't the second one identical to the first one? Or maybe it's just $2^n$ ways to choose students to offer jobs at the bank?
combinatorics discrete-mathematics
combinatorics discrete-mathematics
asked Mar 27 at 4:59
usukidollusukidoll
1,1761033
asked Mar 27 at 4:59
usukidollusukidoll
1,1761033
edited Mar 27 at 5:16
usukidoll
asked Mar 27 at 4:59
usukidollusukidoll
1,1761033
asked Mar 27 at 4:59
usukidollusukidoll
1,1761033
asked Mar 27 at 4:59
usukidollusukidoll
1,1761033
1,1761033
3
$begingroup$
You think too complicatedly. They make a decision for every student they interview. They do $n$ interviews. So they make $n$ decisions.
$endgroup$
– amsmath
Mar 27 at 5:06
$begingroup$
I just realized what happened. For i, the bank can either accept or reject a student after the interview so they have $2^n$ decisions for ii, the student can decision whether or not to accept or decline the bank's job offer, so it's also $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:22
$begingroup$
typos... I meant for ii, the student can decide whether or not to accept or decline the bank's job offer so it's $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:34
add a comment |
3
$begingroup$
You think too complicatedly. They make a decision for every student they interview. They do $n$ interviews. So they make $n$ decisions.
$endgroup$
– amsmath
Mar 27 at 5:06
$begingroup$
I just realized what happened. For i, the bank can either accept or reject a student after the interview so they have $2^n$ decisions for ii, the student can decision whether or not to accept or decline the bank's job offer, so it's also $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:22
$begingroup$
typos... I meant for ii, the student can decide whether or not to accept or decline the bank's job offer so it's $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:34
3
3
$begingroup$
You think too complicatedly. They make a decision for every student they interview. They do $n$ interviews. So they make $n$ decisions.
$endgroup$
– amsmath
Mar 27 at 5:06
$begingroup$
You think too complicatedly. They make a decision for every student they interview. They do $n$ interviews. So they make $n$ decisions.
$endgroup$
– amsmath
Mar 27 at 5:06
$begingroup$
I just realized what happened. For i, the bank can either accept or reject a student after the interview so they have $2^n$ decisions for ii, the student can decision whether or not to accept or decline the bank's job offer, so it's also $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:22
$begingroup$
I just realized what happened. For i, the bank can either accept or reject a student after the interview so they have $2^n$ decisions for ii, the student can decision whether or not to accept or decline the bank's job offer, so it's also $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:22
$begingroup$
typos... I meant for ii, the student can decide whether or not to accept or decline the bank's job offer so it's $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:34
$begingroup$
typos... I meant for ii, the student can decide whether or not to accept or decline the bank's job offer so it's $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I believe that this is overthinking it a bit.
i. There are n students, and each student is a yes or no, leaving n decisions
ii. Every student has 2 options (yes or no), leaving $2^n$
answered Mar 27 at 5:08
Eric LeeEric Lee
802317
$endgroup$
1
$begingroup$
Or $sum_k=0^nbinom nk = (1+1)^n = 2^n$.
$endgroup$
– amsmath
Mar 27 at 5:10
$begingroup$
I got so confused because it is just asking How many decisions... oh wait I think for both it's just $2^n$ because the bank can make a yes or no decision and the student can either make a yes or no decision as well.
$endgroup$
– usukidoll
Mar 27 at 5:18
add a comment |
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$begingroup$
I believe that this is overthinking it a bit.
i. There are n students, and each student is a yes or no, leaving n decisions
ii. Every student has 2 options (yes or no), leaving $2^n$
answered Mar 27 at 5:08
Eric LeeEric Lee
802317
$endgroup$
1
$begingroup$
Or $sum_k=0^nbinom nk = (1+1)^n = 2^n$.
$endgroup$
– amsmath
Mar 27 at 5:10
$begingroup$
I got so confused because it is just asking How many decisions... oh wait I think for both it's just $2^n$ because the bank can make a yes or no decision and the student can either make a yes or no decision as well.
$endgroup$
– usukidoll
Mar 27 at 5:18
add a comment |
$begingroup$
I believe that this is overthinking it a bit.
i. There are n students, and each student is a yes or no, leaving n decisions
ii. Every student has 2 options (yes or no), leaving $2^n$
answered Mar 27 at 5:08
Eric LeeEric Lee
802317
$endgroup$
1
$begingroup$
Or $sum_k=0^nbinom nk = (1+1)^n = 2^n$.
$endgroup$
– amsmath
Mar 27 at 5:10
$begingroup$
I got so confused because it is just asking How many decisions... oh wait I think for both it's just $2^n$ because the bank can make a yes or no decision and the student can either make a yes or no decision as well.
$endgroup$
– usukidoll
Mar 27 at 5:18
add a comment |
$begingroup$
I believe that this is overthinking it a bit.
i. There are n students, and each student is a yes or no, leaving n decisions
ii. Every student has 2 options (yes or no), leaving $2^n$
answered Mar 27 at 5:08
Eric LeeEric Lee
802317
$endgroup$
I believe that this is overthinking it a bit.
i. There are n students, and each student is a yes or no, leaving n decisions
ii. Every student has 2 options (yes or no), leaving $2^n$
answered Mar 27 at 5:08
Eric LeeEric Lee
802317
answered Mar 27 at 5:08
Eric LeeEric Lee
802317
answered Mar 27 at 5:08
Eric LeeEric Lee
802317
answered Mar 27 at 5:08
Eric LeeEric Lee
802317
802317
1
$begingroup$
Or $sum_k=0^nbinom nk = (1+1)^n = 2^n$.
$endgroup$
– amsmath
Mar 27 at 5:10
$begingroup$
I got so confused because it is just asking How many decisions... oh wait I think for both it's just $2^n$ because the bank can make a yes or no decision and the student can either make a yes or no decision as well.
$endgroup$
– usukidoll
Mar 27 at 5:18
add a comment |
1
$begingroup$
Or $sum_k=0^nbinom nk = (1+1)^n = 2^n$.
$endgroup$
– amsmath
Mar 27 at 5:10
$begingroup$
I got so confused because it is just asking How many decisions... oh wait I think for both it's just $2^n$ because the bank can make a yes or no decision and the student can either make a yes or no decision as well.
$endgroup$
– usukidoll
Mar 27 at 5:18
1
1
$begingroup$
Or $sum_k=0^nbinom nk = (1+1)^n = 2^n$.
$endgroup$
– amsmath
Mar 27 at 5:10
$begingroup$
Or $sum_k=0^nbinom nk = (1+1)^n = 2^n$.
$endgroup$
– amsmath
Mar 27 at 5:10
$begingroup$
I got so confused because it is just asking How many decisions... oh wait I think for both it's just $2^n$ because the bank can make a yes or no decision and the student can either make a yes or no decision as well.
$endgroup$
– usukidoll
Mar 27 at 5:18
$begingroup$
I got so confused because it is just asking How many decisions... oh wait I think for both it's just $2^n$ because the bank can make a yes or no decision and the student can either make a yes or no decision as well.
$endgroup$
– usukidoll
Mar 27 at 5:18
add a comment |
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3
$begingroup$
You think too complicatedly. They make a decision for every student they interview. They do $n$ interviews. So they make $n$ decisions.
$endgroup$
– amsmath
Mar 27 at 5:06
$begingroup$
I just realized what happened. For i, the bank can either accept or reject a student after the interview so they have $2^n$ decisions for ii, the student can decision whether or not to accept or decline the bank's job offer, so it's also $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:22
$begingroup$
typos... I meant for ii, the student can decide whether or not to accept or decline the bank's job offer so it's $2^n$ decisions.
$endgroup$
– usukidoll
Mar 27 at 5:34