How many distinct 8-digit positive integers can be formed by rearranging all of the digits… — GMAT

How many distinct 8-digit positive integers can be formed by rearranging all of the digits $1, 1, 1, 2, 2, 3, 3,$ and $3$ such that the resulting integer is even and no two $1$ s are adjacent?

Cevap: 40

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To ensure the 8-digit number is even, the units (8th) digit must be 2. This leaves 7 positions to fill with three 1s, one 2, and three 3s. By arranging the 4 non-1 digits ({2, 3, 3, 3}) first, there are 4! / (1! 3!) = 4 distinct arrangements. Placing 4 digits creates 5 distinct gaps where the 1s can be placed without being adjacent. Choosing 3 gaps out of 5 for the three identical 1s gives C(5, 3) = 10 ways. Thus, the total number of valid integers is 4 * 10 = 40.

Adım Adım Çözüm

Fix the last digit to satisfy the even integer restriction

The 8th digit is fixed as 2, leaving 7 positions to fill with the remaining digits {1, 1, 1, 2, 3, 3, 3}.

An integer is even if and only if its units digit is even. The digit 2 is the only even digit in the set.

Calculate the arrangements of the non-restricted digits

4! / (1! * 3!) = 4 distinct arrangements.

Arranging the four non-1 digits ({2, 3, 3, 3}) first sets up the framework for placing the restricted 1s.

Place the three identical 1s into the gaps using the combination formula

C(5, 3) = 10 distinct gap selections.

The 4 arranged digits create 5 gaps. Selecting 3 distinct gaps ensures no two 1s are adjacent.

Multiply the arrangements of non-1 digits by the number of gap choices

4 * 10 = 40.

By the Fundamental Counting Principle, each non-1 arrangement can be combined with any valid gap placement.

Anahtar Kavram

Permutations with Indistinguishable Objects and Non-Adjacency Restrictions (Gap Method)

Tahmini Süre:2m 0s

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