NIMCET 2007 — Reasoning PYQ
NIMCET | Reasoning | 2007How many arrangements of four 's, two 's and two 's are there in which the first occurs before the first ?
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How many arrangements of four 's, two 's and two 's are there in which the first occurs before the first ?
We are given a total of elements to arrange:
Number of 's =
Number of 's =
Number of 's =
The total number of unrestricted ways to arrange these identical objects containing groups of identical elements is given by the formula for multinomial permutations:
Substituting our values into the formula:
The problem states that the first must occur before the first .
Let's focus strictly on the sub-positions containing the 's and 's. There are positions taken up by these non-zero elements. By symmetry, because there are an equal number of 's (two) and 's (two), the arrangement of these four digits can only start with either a or a .
Let's look at the first appearing digit among the sub-sequence of 's and 's:
Scenario A: The first digit in this sub-sequence is a (This satisfies the condition that the first occurs before the first ).
Scenario B: The first digit in this sub-sequence is a (This violates the condition).
Because the counts of 's and 's are perfectly symmetrical, exactly half of the total valid permutations will have the first appearing before the first , and the other half will have the first appearing before the first .
Therefore, the number of favorable arrangements is exactly half of the total unrestricted arrangements: