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tiny-mole [99]
3 years ago
7

How many different letter permutations, of any length, can be made using the letters M O T T O (For instance, there are 3 possib

le permutations of length 1.)
Mathematics
1 answer:
Tatiana [17]3 years ago
5 0

Answer:

89

Step-by-step explanation:

Given that

2 O, 2 T and 1 M

Now based on this, the following ways are there  

1

Three ways i.e. {M,T,O}

2

XX or XY

XX in 2C1 = two ways i.e. {OO or TT}

XY in 3C2 × 2! = six ways

3

XXY or XYZ

XXY in 2C1 × 2C1 × 3! ÷ 2! = twelve ways

XYZ in 3C3 × 3! = six ways

4

XXYY or XXYZ

XXYY = 4! ÷ (2! × 2!) = six ways

XXYZ in 2C1 ×  4! ÷ 2! = twenty four ways

5

= 5! ÷ (2! ×2!)

= 120 ÷ 4

= 30

Therefore, the total is

= 3 + (2 +6)+ (12 +6) + (6 +24) + 30

= 89

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Find total number of integers.

a_1=10,a_n=60,d=1 \\a_n=a_1+(n-1)d \\60=10+(n-1)1 \\n-1=50 \\n=51

Find how many integers is divisible by 2.

a_1=10,a_n=60,d=2
\\a_n=a_1+(n-1)d
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Eliminate even numbers.

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This array contains 51 - 26 = 25 numbers.

Eliminate numbers before the first number divisible by 3 and after the last number divisible by 3.

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This array contains 25 - 3 = 22 numbers.

Now we should eliminate numbers divisible by 3: 15, 21, 27...

a_1=15,a_n=57,d=6,n=?
\\a_n=a_1+(n-1)d
\\57=15+(n-1)6
\\6(n-1)=42
\\n-1=7
\\n=8

There are 8 such numbers.

Therefore, there are 25 - 8 = 17 numbers that <span>can be evenly divided by neither 2 nor 3</span>

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