some other permutations and combinations questions

Q1) In a conference 10 speakers are present. If S₁ wants to speak before S₂ & S₂ wants to speak after S₃, then number of ways all the 10 speakers can give their speeches with the above restriction if the remaining 7 speakers have no objection to speak at any number is?

Q2) Passengers are to travel by a double decked bus which can accommodate 13 in the upper deck and 7 in the lower deck. The number of ways that they can be divided if 5 refuse to sit in the upper deck and 8 refuse to sit in the lower deck.

What I have done is:-
A1) S₁S₃S₂||||||||
S₃S₁S₂ ||||||||
Now we can arrange them acc. to :-
10C3 *2! * 10C7 * 7!

but the answer is 10!/3.

A2) 13C8* 8!* 7C5 * 5!
but the answer is 21.

please help

dknight said:

Q1) In a conference 10 speakers are present. If S₁ wants to speak before S₂ & S₂ wants to speak after S₃, then number of ways all the 10 speakers can give their speeches with the above restriction if the remaining 7 speakers have no objection to speak at any number is?

Q2) Passengers are to travel by a double decked bus which can accommodate 13 in the upper deck and 7 in the lower deck. The number of ways that they can be divided if 5 refuse to sit in the upper deck and 8 refuse to sit in the lower deck.

What I have done is:-
A1) S₁S₃S₂||||||||
S₃S₁S₂ ||||||||
Now we can arrange them acc. to :-
10C3 *2! * 10C7 * 7!

but the answer is 10!/3.

Click to expand...

There are three class of speakers: S₁ and S₂, S₃, and everybody else. Once you count how many ways those three speakers can be arranged within the ten positions (such that s₁ and s₂ both precede s₃), you have to multiply by the permutations of the remaining seven speakers in seven positions.

EDIT. Looking at the book answer, perhaps there is an easier way. Find the permutations of all ten speakers, and then multiply by the fractional probability that s₃ does come after both s₁ and s₂.

A2) 13C8* 8!* 7C5 * 5!
but the answer is 21.

please help

Click to expand...

It doesn't say,but I presume the total number of passengers is equal to the total number of seats (20). With 5 grabbing seats on the lower level and 8 headed for the top, you are left with 2 lower-level and 5 upper-level seats. The seats are not assigned .. just "upper" or "lower" .. so use Combinations to apportion the remaining 7 people.

Let us see your next attempt, using these hints!

dknight said:

Q1) In a conference 10 speakers are present. If S₁ wants to speak before S₂ & S₂ wants to speak after S₃, then number of ways all the 10 speakers can give their speeches with the above restriction if the remaining 7 speakers have no objection to speak at any number is?
What I have done is:-
A1) S₁S₃S₂||||||||
S₃S₁S₂ ||||||||
Now we can arrange them acc. to :-
10C3 *2! * 10C7 * 7!

but the answer is 10!/3.

Click to expand...

Actually the book's answer is incorrect. The correct answer is \(\displaystyle \dfrac{{10!}}{{3!}}\).

To see that suppose we have the string \(\displaystyle ABCDEFGXXX\).
There are \(\displaystyle \dfrac{{10!}}{{3!}}\) ways to rearrange that string.
In each of those rearrangements, there is only one way to replace the X's with \(\displaystyle S_1,~S_2,~S_3\) in that order.