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3 years ago

How many distinct arrangements can you make using the letters in the word EXPERIMENT

Mathematics

2 answers:

MissTica3 years ago

8 0

You can make 604,800 distinct arrangements in the word EXPERIMENT (option c). I hope this helps!

Ksju [112]3 years ago

7 0

The correct answer is c

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The product of 2 and x

Elis [28]

Answer:

Step-by-step explanation:

two times x is 2x

when multiplying anything

it is 1x*2

which is 2*1=2

2*x

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4 years ago

Let the universal set U = {q, r, s, t, u, v, w, x,

Karo-lina-s [1.5K]

AUB means adding all the members of set A and all the members of set B
A=(q,s,u,w,y) + set B=(q,s,u,y,s)
Therefore,AUB =(q,s,u,w,y,z)

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3 years ago

A bag of marbles contains 6 blue marbles, 2 yellow marbles, 4 red marbles, and 1 green marble. What is the probability of reachi

Mice21 [21]

Total marbles = 13

Number of yellow = 2

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3 years ago

b) How many ways can you deal cards (from a deck of 52) to 4 people when each player gets 7 cards. 2 hidden and 5 visible. Assum

Mariana [72]

Answer:

${52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4$

Step-by-step explanation:

The dealing of the cards can be seen in the following way:

We first have to choose which 7 cards are going to be dealt to the 1st player. So we have to pick 7 cards out of the 52 available cards. This can be done in ${52 \choose 7}$ ways. Now that we have chosen which 7 cards are going to be dealt to the 1st player, we have to choose which 2 of them are going to be the hidden ones. So we have to pick 2 cards out of the 7 cards to be the hidden ones. This can be done in ${7 \choose 2}$ ways. At this point we now know which cards are being dealt to the 1st player, and which ones are hidden for him. Then we have to choose which 7 cards to deal to the 2nd player, out of the remaining ones. So we have to pick 7 out of 52-7=45. (since 7 have been already dealt to the 1st player). This can be done in ${45 \choose 7}$ . Then again, we have to pick which 2 are going to be the hidden cards for this 2nd player. So we have to pick 2 out of the 7. This can be done in ${7 \choose 2}$ ways. Then we continue with the 3rd player. We have to choose 7 cards out of the remaining ones, which at this point are 52-7-7=38. This can be done in ${38 \choose 7}$ . And again, we have to choose which ones are the hidden ones, which can be chosen in ${7 \choose 2}$ ways. Finally, for the last player, we choose 7 out of the remaining cards, which are 52-7-7-7=31. This can be done in ${31 \choose 7}$ ways. And choosing which ones are the hidden ones for this player can be done in ${7 \choose 2}$ ways. At the end, we should multiply all our available choices on each step, to get the total choices or total ways to deal the cards to our 4 players (since dealing the cards is a process of several steps with many choices on each step).

${52 \choose 7}{7 \choose 2}{45 \choose 7}{7 \choose 2}{38 \choose 7}{7 \choose 2}{31 \choose 7}{7 \choose 2}={52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4$

7 0

4 years ago

Alex bought a 6-pack of sports drink bottles that each had a volume of 350 ml.

Kaylis [27]

Answer:

1,050ml

Step-by-step explanation:

6 x 350= 2,100

3 x 350= 1,050

2,100 - 1,050 = 1050ml

7 0

3 years ago

Read 2 more answers