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How Many 6-digit Telephone Numbers Can Be Constructed with Digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 If Each Number Starts with 35 and No Digit Appears More than Once?

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प्रश्न

How many 6-digit telephone numbers can be constructed with digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each number starts with 35 and no digit appears more than once?

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उत्तर

Total available digits = 10
Out of these, 3 and 5 have already been used to make the first two digits.
∴ Number of available digits = 8
The telephone number consists of 6 digits. The initial numbers have already been fixed as 35.

Since repetition is not allowed, the number of telephone numbers that can be formed is equal to the number of  arrangements of the 8 digits, taken 4 at a time.
⇒ 8P4 =\[\frac{8!}{4!} = 8 \times 7 \times 6 \times 5 = 1680\]

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Factorial N (N!) Permutations and Combinations
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 16: Permutations - Exercise 16.3 [पृष्ठ २९]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 16 Permutations
Exercise 16.3 | Q 26 | पृष्ठ २९

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