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shivusuja

Member

Registered: 2006-09-14

Posts: 56

Permutation And Combination

PLEASE HELP ME OUT OF THIS TWO SUMS

1: In how many ways can the letter of the word "PERMUTATIONS" be arranged if there are always 4 letters between P & s ?

2: In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsary for every student?

THANK YOU.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Permutation And Combination

1 Do you mean exactly 4 letters? In this case …

If P occurs in the 1st, 2nd, 3rd, 4th or 5th place, there is only one place for the S to be, namely 5 places to the right. Similarly if P occurs in the 8th, 9th, 10th, 11th or 12th place, there is only one place for the S to be, namely 5 places to the left. If P occurs in the middle two places, the S can be either 5 places to the left or 5 places to the right of the P. Hence there are 5 + 5 + 2×2 = 14 ways for the P and the S to take the first two places.

Then the other 10 letters can be placed anywhere, namely in (10!)⁄(2!) (dividing by 2! because of the repeated T) = 1814400 ways. So the total number of ways is 14 × 1814400 = 25401600.

If there are to be at least 4 letters …

If P occurs in the 1st place, S can be in the 6th–12th place (7 places).
If P occurs in the 2nd place, S can be in the 7th–12th place (6 places).
.
.
.
If P occurs in the 5th place, S can be in the 10th–12th place (3 places).
If P occurs in the 6th place, S can be in the 1st, 11th or 12th place (3 places).
If P occurs in the 7th place, S can be in the 1st, 2nd or 12th place (3 places).
If P occurs in the 8th place, S can be in the 1st–3rd places (3 places).
.
.
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If P occurs in the 12th place, S can be in the 1st–7th place (7 places).

So this time the number of ways for P and S to pick the first two spots is (adding up all the above places) 56. And the total number of permutations this time is 56 × 1814400 = 101606400.

I’ve given two answers above, depending on whether you mean exactly 4 letters between P and S or at least 4 letters between P and S.

2 The student effectively has to choose 3 courses from 7 courses available after taking out the 2 compulsory courses. Hence [sup]7[/sup]C[sub]3[/sub] = 35.

Last edited by JaneFairfax (2007-09-21 02:40:22)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Permutation And Combination

(1) Ignore the S for the moment. Now we've just got PERMUTATION and want to know how many ways there are of arranging that. If all the letters were different then it would be 11!, but because there are two T's, we need to divide that value by two to avoid repetition.

Now let's think about the S. If the P is 5th or further, then the S can be placed left of it.
If the P is 7th or earlier, then the S can be placed right of it.
That means that there are two places for the S if the P is 5th, 6th or 7th, and one otherwise.

3/11 of the arrangements will have the P in one of those places, so the final answer is (3/11x2 + 8/11x1)(11!/2) = 25,401,600.

(2) If two of the courses are compulsory, then really the student is just picking 3 subjects out of 7.

Therefore, there are 7C3 = 35 combinations of subjects.

Edit: Bah, Jane beat me. Still, at least we confirm each other.

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Why did the vector cross the road?
It wanted to be normal.