Soru

Zorluk: Çok zorPermutations

Calculate the total number of distinct four-digit numbers that can be formed using the digits 1,2,3,4,5,6,1, 2, 3, 4, 5, 6, and 77 without repetition, such that the resulting number is divisible by either 44 or 55.

Cevap: 320

Cevap

The total number of such four-digit numbers is 320320.
The total number of valid four-digit numbers is found by adding the count of numbers divisible by 55 (120120) to the count of numbers divisible by 44 (200200). Since a number ending in 55 is odd, it cannot be divisible by 44, making the two conditions mutually exclusive. Thus, the total count is 120+200=320120 + 200 = 320.

Adım Adım Çözüm

1
Calculate the number of four-digit numbers divisible by 55.
For a number to be divisible by 55, its units digit must be 55 (since 00 is not available). There is 11 choice for the units digit. The remaining 33 positions are filled from the remaining 66 available digits in 6P3=6×5×4=120^{6}P_{3} = 6 \times 5 \times 4 = 120 ways.
Divisibility by 55 requires the last digit to be 55.
2
Calculate the number of four-digit numbers divisible by 44.
A number is divisible by 44 if its last two digits form a multiple of 44. Using distinct digits from {1,2,3,4,5,6,7}\{1, 2, 3, 4, 5, 6, 7\}, the valid two-digit endings are 12,16,24,32,36,52,56,64,72,12, 16, 24, 32, 36, 52, 56, 64, 72, and 7676 (1010 valid pairs). For each pair, the first two positions are filled from the remaining 55 digits in 5P2=5×4=20^{5}P_{2} = 5 \times 4 = 20 ways. Thus, total ways = 10×20=20010 \times 20 = 200.
Divisibility by 44 depends entirely on the last two digits.
3
Check for overlap (numbers divisible by both 44 and 55).
A number divisible by 55 must end in 55, which is an odd digit. All multiples of 44 must end in an even digit. Hence, no number is divisible by both 44 and 55 in this set. The overlap is 00.
The two events are mutually exclusive.
4
Apply the addition principle of counting.
Total = 120+2000=320120 + 200 - 0 = 320.
Add the counts of the two mutually exclusive sets.

Anahtar Kavram

Restricted Permutations and Mutually Exclusive Events
Tahmini Süre:1m 30s
Bu soruyu puanla