The first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ?
A) 1/7
B) 8!
C) 7!
D) 1/14
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Q:
The first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ?
A) 1/7
B) 8!
C) 7!
D) 1/14
Answer: D) 1/14
Read Description:
The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14
Hence the answer is 1/14.
The first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ?
A) 1/7
B) 8!
C) 7!
D) 1/14
Answer: D) 1/14
Read Description:
The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14
Hence the answer is 1/14.
Answered by
1
Heyy mate ❤✌✌❤
Here's your Answer....
⤵️⤵️⤵️⤵️⤵️⤵️⤵️
The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14
Hence the answer is 1/14.
✔✔✔
Here's your Answer....
⤵️⤵️⤵️⤵️⤵️⤵️⤵️
The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14
Hence the answer is 1/14.
✔✔✔
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