12. P (7,r) = 5040. Find rA. 5B. 6C. 7D. 8

Mathematics

2 answers:

aalyn [17]3 years ago

7 0

Answer:

Step-by-step explanation:

P(7,r) = 5040

$P(n,r)=\dfrac{n!}{(n - r)!}\\\\\\n = 7 \\\\\dfrac{7!}{(7 - r)!}=5040\\\\\\7! = (7-r)!*5040\\\\(7-r)!=\dfrac{5040}{7*6*5*4*3*2*1}\\\\\\(7-r)!=\dfrac{5040}{5040}\\\\\\(7-r)!=1\\\\$

(7 - r)! = 1!

7 -r = 1

7 = 1 + r

7 - 1 = r

r = 6

mel-nik [20]3 years ago

4 0

Answer:

12. B or C

13. C

14. A

15. A

Step-by-step explanation:

These problems involve various counting techniques. For combinations and permutations, there are formulas to aid in the counting. For other situations, you may need to list possibilities.

P(7, r) = 5040 . . . . . given

P(7, r) = 7!/(7 -r)! = 5040

5040/(7 -r)! = 5040 . . . . . evaluate 7!

(7 -r)! = 1 . . . . . . . . . . . multiply by (7-r)!/5040

We know that both 0! and 1! = 1, so this has two solutions:

7 -r = 0 ⇒ r = 7

7 -r = 1 ⇒ r = 6

Both choices B and C are correct.

The attached calculator output confirms this result.

The formula of choice C is correct:

C(n, r) = n!/((n-r)!r!)

The number of permutations of the 9 letters is reduced by the numbers of permutations of letters that are indistinguishable. That is, the 9 letters can be rearranged 9! ways, but the 4 Es are indistinguishable, so a factor of 4! must be removed from that result. Similarly, the two Ns and two Ss are indistinguishable, so additional factors of 2! must be removed. The number of arrangements is ...

9!/(4!2!2!) . . . . matches choice A

The 5 students can be arranged in 5 consecutive chairs in P(5, 5) = 5! = 120 ways. Those 5 consecutive chairs can start in chairs 1 through 6, so there are 6×120 = 720 ways the 5 students can sit together in 10 chairs.