Let m (respectively, n ) be the number of 5 -digit integers obtained by using the digits 1 , 2 , 3 , 4 , 5 with repetitions (respectively, without repetitions) such that the sum of any two adjacent digits is odd. Then m n is equal to

Question

Let m (respectively, n ) be the number of 5 -digit integers obtained by using the digits 1 , 2 , 3 , 4 , 5 with repetitions (respectively, without repetitions) such that the sum of any two adjacent digits is odd. Then m n is equal to, 9, 12, 15, 18

MARKS App · Accepted Answer

We have, m is 5 -digits number using digits 1 , 2 , 3 , 4 , 5 with repetition such that sum of two adjacent digit is odd and n is 5 -digits number using digits 1 , 2 , 3 , 4 , 5 without repetitions such that sum of any two adjacent digits is odd. Sum of two digits are odd if one is even and other is odd. Even = 2 , 4 Odd = 1 , 3 , 5 Case I Digit is repeated. Two possibilities (a) odd even odd even odd = 3 × 2 × 3 × 2 × 3 = 108 (b) even odd even odd even = 2 × 3 × 2 × 3 × 2 = 72 ∴ m = 108 + 72 = 180 Case II Digit is not repeated. The possibility of arrangement is odd even odd even odd = 3 × 2 × 2 × 1 × 1 = 12 n = 12 ∴ m n = 180 12 = 15

Search any question & find its solution

Looking for more such questions to practice?