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Nov 2016 p63 q3
2706
Numbers are formed using some or all of the digits 4, 5, 6, 7 with no digit being used more than once.
Show that, using exactly 3 of the digits, there are 12 different odd numbers that can be formed.
Find how many odd numbers altogether can be formed.
Solution
(i) To form a 3-digit odd number using the digits 4, 5, 6, 7, the last digit must be odd. The odd digits available are 5 and 7.
For numbers ending in 5, the remaining two digits can be chosen from 4, 6, 7. This can be done in \(^3P_2 = 6\) ways.
For numbers ending in 7, the remaining two digits can be chosen from 4, 5, 6. This can also be done in \(^3P_2 = 6\) ways.
Therefore, the total number of 3-digit odd numbers is \(6 + 6 = 12\).
(ii) Consider numbers with different numbers of digits:
1-digit numbers: The odd digits are 5 and 7, so there are 2 numbers.
2-digit numbers: The last digit must be odd (5 or 7). For each choice of the last digit, the first digit can be chosen from the remaining 3 digits. This gives \(^3P_1 \times 2 = 6\) numbers.
3-digit numbers: As calculated in part (i), there are 12 numbers.
4-digit numbers: The last digit must be odd (5 or 7). For each choice of the last digit, the first three digits can be chosen from the remaining 3 digits. This gives \(^3P_3 \times 2 = 12\) numbers.
Therefore, the total number of odd numbers is \(2 + 6 + 12 + 12 = 32\).
