The letters of the word ‘CLIFTON’ are placed at random in a row. W

1.	The letters of the word ‘CLIFTON’ are placed at random in a row. What is the chance that two vowels come together?
Answer» given: word “CLIFTON” Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) In the random arrangement of the alphabets of word “CLIFTON” we have to find the probability that vowels come together total possible outcomes of arranging the alphabets are 7! therefore n(S)=7! let E be the event that vowels come together number of vowels in SOCIAL is I, O therefore, number of ways to arrange them so (I, O) come together n(E)= 6! × 2! P(E) = \(\frac{n(E)}{n(S)}\) P(E) = \(\frac{6!2!}{7!}\) = \(\frac{2}{7}\)

The letters of the word ‘CLIFTON’ are placed at random in a row. What is the chance that two vowels come together?

Answer»

given: word “CLIFTON”

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\)

In the random arrangement of the alphabets of word “CLIFTON” we have to find the probability that vowels come together

total possible outcomes of arranging the alphabets are 7!

therefore n(S)=7!

let E be the event that vowels come together

number of vowels in SOCIAL is I, O

therefore, number of ways to arrange them so (I, O) come together

n(E)= 6! × 2!

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{6!2!}{7!}\) = \(\frac{2}{7}\)

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