Let �. Let � be the number of 9-digit numbers formed using the digits of the set � such that only one digit is repeated and it is repeated exactly twice. Let � be the number of 9-digit numbers formed using the digits of the set � such that only two digits are repeated and each of these is repeated exactly twice. Then,

Let S = {1,2,3,4,5,6,7,8,9}. Let x be the number of 9-digit numbers formed such that only one digit is repeated exactly twice and y be the number of 9-digit numbers formed such that only two digits are repeated exactly twice. Then

Q. Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let x be the number of 9-digit numbers formed using the digits of the set S such that only one digit is repeated and it is repeated exactly twice. Let y be the number of 9-digit numbers formed using the digits of the set S such that only two digits are repeated and each of these is repeated exactly twice. Then,

(A) 56x = 9y

(B) 21x = 4y

(D) 14x = 3y

Correct Answer: 21x = 4y

Explanation

First, find x.

Choose the digit which is repeated twice from 9 digits:

= 9C1

Choose the remaining 7 distinct digits from the remaining 8 digits:

= 8C7

Total arrangements of 9 digits where one digit is repeated twice:

x = 9C1 × 8C7 × 9! / 2!

Now find y.

Choose 2 digits which are repeated twice:

= 9C2

Choose remaining 5 distinct digits from remaining 7 digits:

= 7C5

Total arrangements:

y = 9C2 × 7C5 × 9! / (2!)²

Now take ratio:

x / y = (9C1 × 8C7 × 2!) / (9C2 × 7C5)

Simplifying,

x / y = 4 / 21

⇒ 21x = 4y

Hence, the correct relation is 21x = 4y.

Explanation

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