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Exercise 6.2 · Q14

Q.How many numbers having 5 digits can be formed with the digits 0, 2, 3, 4 and 5 if repetition of digits is not allowed. How many of these are divisible by 5?

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Using digits 0,2,3,4,50,2,3,4,5 without repetition, there are 9696 valid 5-digit numbers (leading digit ≠0\neq 0), of which 4242 are divisible by 5 (units digit 00 or 55).

[!FORMULA] By the multiplication principle with the restriction that the leading digit cannot be 00: total 5-digit numbers == (choices for the first digit, excluding 00) ×\times (permutations of the remaining 4 digits in the remaining 4 places) =4×4!=4\times4!. For divisibility by 5, the units digit must be 00 or 55; each case is counted separately (with the first-digit restriction reapplied) and the two cases summed.

  1. Total 5-digit numbers (no repetition), digits available {0,2,3,4,5}\{0,2,3,4,5\}: the first (leftmost) digit cannot be 00, so it has 44 choices (2,3,4,52,3,4,5). The remaining 4 positions are filled by arranging the remaining 4 digits (which now include 00) in 4!4! ways.

  2. Compute: 4×4!=4×24=964\times4!=4\times24=96.

  3. Divisible by 5 — Case A: units digit =0=0. The units place is fixed at 00. The first digit can be any of the remaining 4 nonzero digits {2,3,4,5}\{2,3,4,5\}: 44 choices. The middle 3 digits are filled by arranging the remaining 3 digits in 3!3! ways.

  4. Case A count: 4×3!=4×6=244\times3!=4\times6=24.

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