SOLUTION: I just need someone to kindly look over these problems and let me know if I did them correctly.
1. In how many ways can three drivers be hired from a pool of 10 applicants to dr
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Question 462120: I just need someone to kindly look over these problems and let me know if I did them correctly.
1. In how many ways can three drivers be hired from a pool of 10 applicants to drive three identical cars?
3!10?
2. In how many ways can three drivers be hired from a pool of 10 applicants to drive a sedan, a minivan, and a pickup truck?
3!10 x 3!?
3. A coin is flipped 12 times and the sequence of heads and tails is observed. In how many ways can the sequence consist of exactly 7 heads and 5 tails?
I dont know how to set this one up!
5. In how many ways can 2 C’s, 4 D’s, and 6 F’s be awarded to 12 politicians, one letter for each politician?
2!4!6!12!?
6. 15 people enter an art contest. In how many different ways can a champion, a first runner-up, and a second runner-up be selected?
15!3
7. In how many different ways can a committee of 9 members be chosen from 6 freshmen, 5 sophomores, and 4 juniors if the committee must consist of 4 freshmen, 3 sophomores, and 2 juniors?
9!6!5!4! x 4!3!2!
8.In how many different ways can a committee of 9 members be chosen from 6 freshmen, 5 sophomores, and 4 juniors if the committee must consist of at least one freshman?
9!6!5!4! x 6!
.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I just need someone to kindly look over these problems and let me know if I did them correctly.
1. In how many ways can three drivers be hired from a pool of 10 applicants to drive three identical cars?
Ans: 10C3 = (10*9*8)/(1*2*3) = 120 ways because you are counting groups of 3.
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2. In how many ways can three drivers be hired from a pool of 10 applicants to drive a sedan, a minivan, and a pickup truck?
Ans: 10P3 = 10*9*8 = 720 ways because order is implied.
-----------------------------------
3. A coin is flipped 12 times and the sequence of heads and tails is observed. In how many ways can the sequence consist of exactly 7 heads and 5 tails?
------
Ans: 12C5 = (12*11*10*9*8)/(1*2*3*4*5) = 792
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5. In how many ways can 2 C’s, 4 D’s, and 6 F’s be awarded to 12 politicians, one letter for each politician?
Ans: 12!/(2!*4!*6!) = 13,860 ways
Similar to # of 12-letter words with 2C's, 4D's, and 6F's.
--------------------------------------
6. 15 people enter an art contest. In how many different ways can a champion, a first runner-up, and a second runner-up be selected?
Ans: Order is implied: 15P3 = 15*14*13 = 2730 ways
-----------------------------------------------------
7. In how many different ways can a committee of 9 members be chosen
from 6 freshmen, 5 sophomores, and 4 juniors if the committee must consist of
4 freshmen, 3 sophomores, and 2 juniors?
Ans: 6C4*5C3*4C2 = 15*10*6 = 900 ways
------------------------------------------------
8.In how many different ways can a committee of 9 members be chosen from
6 freshmen, 5 sophomores, and 4 juniors if the committee must consist of at least one freshman?
Ans:
#'s in the order of (frosh;soph;juniors)
The limits are 6 frosh;5 soph;4 juniors
===============================================
1:5;3 (this means 1 frosh; 5 soph; 3 juniors): # of ways = 6*5*4C3 = 6*5*4 = 120
1;4;4::: # of ways = 6*5C4*4C4 = 30
--------
I'll leave the rest for you to figure out.
===================
2;4;3
2;3;4
-------
3:4;2
3:3:3
3;2:4
----------
4:4:1
4;3;2
4;2;3
4;1;4
--------
5;4;0
5;3;1
5;2;2
5;1;3
5;0;4
-------
6;3;0
6;2;1
6;1;2
6;0;3
==============
Cheers,
Stan H.
================
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