Question

There are 20 identical sticks lined up in a row occupying 20 distinct places as follows: 11111111111111111111· Six of them are to be chosen. (a) How many choices are there? (b) How many choices are there if no two of the chosen sticks can be consecutive? (c) How many choices are there if there must be at least two sticks between each pair of chosen sticks?

Answer 1

Answer:

a)38760

b)5005

c)210

Given that

20 identical sticks

We have to 6 of them

a)

The number of choice will be .

b)

The number of choices if no two of the chosen sticks can be consecutive will be .

c)

|xx|xx|xx|xx|xx|

Total seven positions are available

So

4-0-0-0-0-0-0 7!/6!=7

3-1-0-0-0-0-0 7!/5!=42

2-2-0-0-0-0-0 7!/2!5!=21

2-1-1-0-0-0-0 7!/2!4!=105

1-1-1-1-0-0-0 7!/4!3!=35

Now by adding all 7+42+21+105+35=210

Explanation:

Answer 2

a)38760

b)5005

c)210

Explanation:

Given that

20 identical sticks

We have to 6 of them

a)

The number of choice will be
`_(6)^(20)\textrm{c}`.

`_(6)^(20)\textrm{c}=(20!)/(6!.14!)`

`_(6)^(20)\textrm{c}=38760`

b)

The number of choices if no two of the chosen sticks can be consecutive will be
`_(6)^(15)\textrm{c}`.

`_(6)^(15)\textrm{c}=(15!)/(6!9!)`

`_(6)^(15)\textrm{c}=5005`

c)

|xx|xx|xx|xx|xx|

Total seven positions are available

So

4-0-0-0-0-0-0 7!/6!=7

3-1-0-0-0-0-0 7!/5!=42

2-2-0-0-0-0-0 7!/2!5!=21

2-1-1-0-0-0-0 7!/2!4!=105

1-1-1-1-0-0-0 7!/4!3!=35

Now by adding all 7+42+21+105+35=210