Mathematics, 14.09.2019 03:10 batmannn1516

There are 14 juniors and 16 seniors in a chess club. a) from the 30 members, how many ways are there to arrange 5 members of the club in a line? b) how many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line? 0 c) if the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there? d) if the club sends either 4 juniors or 4 seniors, how many possible groupings are there?

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Answer from: Angelanova69134

No. of juniors = 14

No. of seniors = 16

Total students = 30

A) From the 30 members, how many ways are there to arrange 5 members of the club in a line?

Since we are asked about arrangement so we will use permutation

Formula : $^nP_r=\frac{n!}{(n-r)!}$

n = 30

r = 5

$^{30}P_5=\frac{30!}{(30-5)!}$

$^{30}P_5=17100720$

So, From the 30 members, there are 17100720 ways to arrange 5 members of the club in a line?

B) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?

Out of 16 seniors 2 will be selected

So, 3 places are vacant

Remaining students = 30-2 = 28

So, out of 28 students 3 students will be selected

No. of ways = $^{16}P_2 \times ^{28}P_3$

No. of ways = $\frac{16!}{(16-2)!}\times\frac{28!}{(28-3)!}$

= 4717440

There are 4717440 ways to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line.

C)If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?

Since we are not asked about arrangement so we will use combination

Out of 16 seniors 2 will be selected

Out of 14 juniors 2 will be selected

Formula : $^nC_r=\frac{n!}{r!(n-r)!}$

So, No. of possible groupings = $^{16}C_2 \times ^{14}C_2$

= $\frac{16!}{2!(16-2)!} \times \frac{14!}{2!(14-2)!}$

= 10920

If the club sends 2 juniors and 2 seniors to the tournament, there are 10920 possible groupings

D) If the club sends either 4 juniors or 4 seniors, how many possible groupings are there?

Out of 16 seniors 4 will be selected

Out of 14 juniors 4 will be selected

So, No. of possible groupings = $^{16}C_4 + ^{14}C_4$

= $\frac{16!}{4!(16-4)!} + \frac{14!}{4!(14-4)!}$

= 2821

So,If the club sends either 4 juniors or 4 seniors, there are 2821 possible groupings .

Answer from: Quest

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Answer from: Quest

step-by-step explanation:

given

y + 3 = - 4(x - 2) ← distribute the parenthesis

y + 3 = - 4x + 8 ( subtract 3 from both sides )

y = - 4x + 5 → d

Which of the following equations is equivalent to y+3=-4(x-2)?

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