Total number of six-digit numbers in which only and all the five digits 1, 3, 5, 7, and 9 appear, is

Total number of six-digit numbers in which only and all the five digits 1, 3, 5, 7, and 9 appear, is

(a) 56

(b) (½)(6!)

(c) 6!

(d) (5/2) 6!

Solutions

Expert Solution

Selecting all 5 digits = ⁵C5 = 1 way

We need to select one more digit to make it a 6 digit number = ⁵C1 = 5 ways

Total number of permutations = 6!/2!

Therefore now we have,

Total numbers = ⁵C5×⁵C1×(6!/2!)

= (5/2) 6!

Hence option d is the answer.

Total number is (5/2)6!

Nikhil Thakur answered 1 year ago

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