Question

A disc jokey has 10 songs to play. Four are slow songs, and six are fast songs. Each song is to be played only once. In how many ways can the disc jokey play the 10 songs if
:The songs can be played in any order
:The first song must be a slow song and the last song must be a slow song.
:The first two songs must be fast songs.

Answer 1

10 songs in a sequence can be played in 10! ways

Answer 2

For the second q-
1) you fix first and last as slow songs : 4 * 3
2) rest of the 8 songs can be arranged in 8! ways

so total ways he can play by having first & last as slow songs = 4*3*8!

Answer 3

does that help ?
last q can u try.. .

Answer 4

i donno thats kinda confusing to me...

Answer 5

did u get my method... how i worked out first two ?

Answer 6

can u explain it a bit more how u did that?

Answer 7

10 different songs, so
think 10 spaces to fill with songs ok

_ _ _ _ _ _ _ _ _ _

Answer 8

k

Answer 9

now lets work out how many different ways you can play these 10 songs

Answer 10

First song can be played by choosing any of the 10 songs.
Second song can be played by choosing any of the remaining 9 songs.

(its not 10 becos, repitition is not allowed, so we cant play the same song. one less)

Answer 11

ok

Answer 12

similarly,
Third song can be played in 8 ways
Fourth song can be played in 7 ways

......

Ninth song can be played in 2 ways
Tenth song can be played in 1 ways


total ways = 10 * 9 * 8 * 7 * ...... 2*1 = 10!

Answer 13

ok

Answer 14

above is the fundamental thing u should understand. if u understand this, you can do ANY permutation problem!

Answer 15

did u get the method ?
second q- is little different but you can use the same method there also

Answer 16

i think so

Answer 17

great !

2nd q :
:The first song must be a slow song and the last song must be a slow song.

Answer 18

if you forget about first and last spaces for a while. middle 8 spaces can be arranged in how many ways ?

Answer 19

8! right ?

Answer 20

o ya duh

Answer 21

first and last you fixing with two slow songs. so remaining 8 middle spaces can be arranged in 8! ways. its same logic we used in the first q

S _ _ _ _ _ _ _ _ S

Answer 22

we are not done yet. we have 4 slow songs.
so first & last space can be chosen in 4*3 ways

so total ways = 4*3*8!

Answer 23

makes sense ? let me know if it doesnt... third q is same as 2nd one

Answer 24

o ya ok

Answer 25

:The first two songs must be fast songs.

Answer 26

F F _ _ _ _ _ _ _ _

Answer 27

remaining 8 spaces can be filled in how many ways can u tell me

Answer 28

8 are slow

Answer 29

hmm read the question once. there are only 4 slow songs :P

Answer 30

you simply fix first two as Fast. remaining 8 spaces can be filled in how many ways ?

Answer 31

3 ways

Answer 32

nope. 8 spaces & 8 songs. so its 8! ways right ?

Answer 33

8 ways

Answer 34

yes i already said that

Answer 35

since there are 6 Fast songs,

first two Fast songs can be selected from them in : 6*5 ways.

so total ways = 6*5*8!

Answer 36

its not 8 ways. its 8! ways = 8*7*6*... 2*1

Answer 37

8 \(\ne \) 8!

Answer 38

ok

Answer 39

why do you have exclamation marks after 8?

Answer 40

! is a symbol for factorial

4! = 4*3*3*1

3! = 3*2*1

8! = 8*7*6*5*4*3*2*1

Answer 41

o ok

Answer 42

100! = 100*99*98...... 3*2*1

it just says, to perform series of multiplications as above

Answer 43

for the 3rd q-
we got answer as : 6*5*8!

can you do the factorial and and see what value u get.. ?

Answer 44

6= 6*5*4*3*2*1 5 = 5*4*3*2*1 8 = 8*7*6*5*4*3*2*1 so this is what u were doing and to get a vlue?

Answer 45

hmm no

Answer 46

notice that exclamation sigh is there only for 8

Answer 47

6*5*8! = 30*8! = 30 * 8*7*6*5*4*3*2*1

Answer 48

o 5 * 6 = 30 * 8! = 30 * 8 * 7* 6*5*4*3*2*1 ok