In a four digit number, first digit is greatest and last digit is smallest. Sum of middle two digits is even. How many such four digit numbers, having all digits different, can be there if first digit is 9?
Solution
Correct Answer: Option C
In a four digit number, first digit is 9 and last digit is smallest among 4 digits. Sum of middle two digits is even.
⇒ Between first and last digits, there is a difference of at least 3 digits.
It may come into mind that a difference of 2 digits can suffice, but that would lead to sum of middle two digits being odd (sum of two consecutive digits). But we need the sum to be even.
Combinations for first and last digits can be (9, 0), (9, 1), (9, 2), (9, 3), (9, 4), (9, 5).
Between 0 and 9, there are 4 odd digits and 4 even digits. Two can be chosen such that their sum is even, if both are odd or both are even. This can be done in 4 × 3 + 4 × 3 = 24 ways.
Similarly, for 1 and 9, there can be 3 × 2 + 4 × 3 = 18 ways.
For 2 and 9, there can be 12 ways. For 3 and 9, there can 8 ways. For 4 and 9, there can be 4 ways. For 5 and 9, there can be 1 way.
∴ Total number of 4 digit numbers possible = 24 + 18 + 12 + 8 + 4 + 1 = 67
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