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Butoxors [25]
3 years ago
7

plssssssssssssssssssssss help Six friends, four boys and two girls, went to a movie theater. They wanted to sit in a way so no g

irl sits on either first or last chair. How many such arrangements are possible?
Mathematics
2 answers:
Airida [17]3 years ago
7 0

Answer:

672 ways

Step-by-step explanation:

The total number of ways the friends were supposed to sit without restrictions would have been = 6!

= 720 possible ways.

If the girs should sit on the first or last sit

The possible arrangement= 2!4!

= 2*24

= 48

But now the two girls should be arranged in such a way that the don't sit on either the first or the last chair.

So the possible way now is= way without restriction - way if the girls are to sit at either first or last sit

= 720-48

= 672

tatyana61 [14]3 years ago
5 0

Answer:

672.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

The total number of possibilities to pick 3 parts out of 15 possible parts is given by the following combination:

n=\frac{15!}{(15-3)!3!}=\frac{15*14*13}{3*2*1}\\n=455\ ways\\

(A) There are only three possibilities for which the inspector finds exactly one nonconforming part (NCC, CNC, CCN). Therefore, the probability is:

P(N=1) = \frac{3}{455}=0.006593 =0.6593\%

(B) There are three possibilities  for which the inspector finds exactly one nonconforming part, three possibilities for two nonconforming parts (NNC, CNN, NCN), and one possibility for all nonconforming parts (NNN). The probability that the inspector finds at least one nonconforming part is:

P(N>0) =P(N=1)+P(N=2)+P(N=3) \\P(N>0) = \frac{3}{455}+ \frac{3}{455}+ \frac{1}{455}\\P(N>0) =0.01538 =1.538\%

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