hi i don't know how to solve this probability questions...can you guys help?
question 1: In how many ways can 9 person be seated in a row so that certain three of them must sit together?
question 2: How many integers between 1000 and 9999 inclusive consist of a) Distinct odd digits b) Distinct digits c)From the number of integers obtained in(b), how many are odd integers?
can you guys help solve me this questions??
On
Hint for first question: fix three of them as one seat (and don't forget to permute those three).
Hint for second question: think of all possibilities you can choose for each digit. For instance, part a). One digit can be 1,3,5,7 or 9 so you have 5 choices. Second must be distinct so you have 4 choices and so on.
On
This is for the first one:
We are given 9 people of which 3 (say A,B,C) must sit together. Now, we can treat these 3 people sitting together as 1 unit. Hence, effectively, there are 7 people (6 people + 1 unit).
These 7 people (6 people + 1 unit) can be seated in $^7P_7$ ways ie $7!$ ways.
Thus we have $7!$ ways of seating $6$ persons and the $1$ unit (comprising of ABC). However, the orders in which A,B,C sit will also make a difference to the permutations of the 9 people.
As an example, $$K,L,M,N,O,P,A,B,C$$ is clearly different from $$K,L,M,N,O,P,C,B,A$$
These 3 people can be seated in $^3P_3=3!=6$ ways. Hence, the total number of ways is $$3!\times 7!$$
i visualize it by first shifting a group of three people down and observing:
XXX______ with 6! ways to place the others.
_XXX_____ with 6! ways to place the others.
__XXX____ with 6! ways to place the others.
$4.___XXX___$ with 6! ways to place the others.
____XXX__ with 6! ways to place the others.
_____XXX_ with 6! ways to place the others.
______XXX with 6! ways to place the others.
so there are 7*6! ways to shift three people down.
but those three people XXX can also be arranged 3! ways (see the comments) so the final answer is:
3!*7*6! = 3!*7!