Permutations and combinations thread (1 Viewer)

[Aysce]

Okay, so rather than me making several threads with different questions, I'm compiling my questions into one place.

17. If 6PR =120, find the value of r.

14. How many even numbers of 4 digits can be formed with the figures 3, 4, 7, 8.
a. If no figure is repeated.
b. If repetitions are allowed

In how many ways can these be done?

Thanks to those who help !

 

[Shadowdude]

17.











And 14 doesn't make any sense. Even numbers seated at a round table?

 

[mnmaa]

Okay, so rather than me making several threads with different questions, I'm compiling my questions into one place.

17. If 6PR =120, find the value of r.

14. How many even numbers of 4 digits can be formed with the figures 3, 4, 7, 8
(a) In a row (done)
(b) At a round table.

In how many ways can these be done?

Thanks to those who help !

17.
6Pr=120
(6!)/(6-r)!=120
(6-r)!=6!/120
(6-r)!=6!
3!=6
therefore r=3

 

[Aysce]

17.











And 14 doesn't make any sense. Even numbers seated at a round table?

Thank you for pointing that out. Fixed.

 

[Shadowdude]

14. How many even numbers of 4 digits can be formed with the figures 3, 4, 7, 8.
a. If no figure is repeated.
b. If repetitions are allowed

a. To form a four-digit even number with no repeats:

1. Pick either 4 or 8 to be the last number, that way, the number formed is even... 2 ways
2. Place the remaining three numbers in the other three spots... 3! ways

Answer: 2*(3!) = 12

b. To form a four-digit even number with repeats allowed:

1. Pick either 4 or 8 to be the last number, so the number is even... 2 ways

Our number is _ _ _ 4 or _ _ _ 8.

2. Select 3, 4, 7 or 8 to be the first digit... 4 ways
3. Select 3, 4, 7 or 8 to be the second digit... 4 ways
4. Select 3, 4, 7 or 8 to be the third digit... 4 ways

Answer: 2*(4^3) = 128.

 

[Carrotsticks]

14. How many even numbers of 4 digits can be formed with the figures 3, 4, 7, 8.
a. If no figure is repeated.
b. If repetitions are allowed

In how many ways can these be done?

a)

1. The last digit must be either 4 or 8, so we have 2.

2. We've used up 1 digit, and we have 3 leftover. So we have 3.

3. We've used up 2 digits, so we have 2 leftoever. So we have 2.

4. There is only 1 remaining digit, so it must go in the last position (which is the first digit, I am working backwards).

5. Hence, the asnwer is 2 x 3 x 2 x 1 = 12

b)


1. The last digit must be either 4 or 8, so we have 2.

2. We've used up 1 digit, but repetition is allowed. So we have 4.

3. We've used up 2 digits, but repetition is allowed. So we have 4

4. There is only 1 remaining digit, but repetition is allowed, so we have 4 again.

5. Hence, the asnwer is 2 x 4 x 4 x 4 = 128.

 

[Aysce]

Appreciate the help guys. This topic is killing me..