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PRACTICE PROBLEMS: Find the answers to the following:
a. A and B must be boys and C and D must be girls? b. two of the officers must be boys and two of the officers
must be girls? ANSWERS:
If we were asked how many different arrangements of the
letters in the word STOP can be made, we would write
We would be correct since all letters are different. If
some of the letters were the same, we would reason as given in the following
problem. EXAMPLE:
How many different arrangements of the
letters in the word ROOM can be made? SOLUTION:
We have two letters alike. If we list the
possible arrangements, using subscripts to make a distinction between the O's,
we have
but we cannot distinguish between the O's; RO_{1}O_{2}M
and RO_{2}O_{1}M would be the same arrangement without the
subscript. Only half as many arrangements are possible without the use of
subscripts (a total of 12 arrangements). This leads to the statement: The number
of arrangements of
n items, where there are k groups of like items of size r_{1},r_{2}, . . . r_{k}, respectively, is
given by
in the previous example n was equal
to 4 and two letters were alike; therefore, we would write
EXAMPLE:
How many arrangements can be made using the
letters in the word ADAPTATION? SOLUTION:
We use _{} where n=10 and r_{1} = 2 (two T's) and r_{2} = 3 (three A's) Then
PRACTICE PROBLEMS: Find the number of possible arrangements of the letters in
the following words: 1. WRITE 2. STRUCTURE 3. HERE 4. MILLIAMPERE 5. TENNESSEE ANSWERS:
Although the previous discussions have been associated
with formulas, problems dealing with combinations and permutations may be
analyzed and solved in a more meaningful way without complete reliance upon the
formulas. EXAMPLE: How many fourdigit numbers can be
formed from the digits 2, 3, 4, 5, 6, and 7 a. without repetition? b. with repetition? SOLUTION: The (a) part of the question is a straightforward
permutation problem, and we reason that we want _{
} The (b) part of the question would become quite
complicated if we tried to use the formulas; therefore, we reason as follows: We desire a fourdigit number and find we have six choices
for the first position; that is, we may use any of the digits 2, 3, 4, 5, 6, or
7 in the first position:
Since we are allowed repetition of numbers, then we have
six choices for the second position. In other words, any of the digits 2, 3, 4,
5, 6, or 7 can be used in the second position:
Continuing this reasoning, we also have six choices each
for the third and fourth positions: (6)(6)(6)(6) Therefore, the total number of fourdigit numbers formed
from the digits 2, 3, 4, 5, 6, and 7 with repetition is
EXAMPLE: Suppose, in the previous example, we were to find
how many threedigit odd numbers could be formed from the given digits without
repetition. SOLUTION: We would be required to start in the units column
because an odd number is determined by the units column digit. Therefore, we
have only three choices for the units position; that is, either 3, 5, or 7: For the ten's position, we have only five choices, since
we are not allowed repetition of numbers: Using the same reasoning of no repetition, we have only
four choices for the hundred's position: (3)(5)(4) Therefore, we can form 3^{.}5^{.}4 = 60 threedigit odd numbers from the digits 2, 3, 4, 5, 6, and 7 without repetition. 
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