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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?










0












$begingroup$


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?










share|cite|improve this question











$endgroup$







  • 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















0












$begingroup$


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?










share|cite|improve this question











$endgroup$







  • 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













0












0








0





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 27 at 5:16







usukidoll

















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












  • 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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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 $2^n$






share|cite|improve this answer









$endgroup$








  • 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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 $2^n$






share|cite|improve this answer









$endgroup$








  • 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















2












$begingroup$

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 $2^n$






share|cite|improve this answer









$endgroup$








  • 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













2












2








2





$begingroup$

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 $2^n$






share|cite|improve this answer









$endgroup$



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 $2^n$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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












  • 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

















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