If a code word is defined to be a sequence of different

Notice that as we are dealing with code words then the order of the letters matters.

# of ways to choose 5 different letters out of given 10 letters when the order of chosen letters matters (as for example code word ABCDE is different from BCDEA) is \(P^5_{10}\);

# of ways to choose 4 different letters out of given 10 letters when the order of chosen letters matters is \(P^4_{10}\);

\(Ratio=\frac{P^5_{10}}{P^4_{10}}=\frac{10!}{5!}*\frac{6!}{10!}=\frac{6}{1}\).

Answer: E.
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Numbers of Options applicable for 5 letter digit -> \(10 * 9 * 8 * 7 * 6\)
( as option pool for first digit is 10, for second 9 because one is removed and so on)
Numbers of Options applicable for 5 letter digit -> \(10 * 9 * 8 * 7\)

Required Ratio -> \((10 * 9 *8 * 7 * 6)/(10 * 9* 8 * 7)\) = \(6:1\)

Number of 5-letter code formed from 10 letters: \(=10*9*8*7*6\)
Number of 4-letter code formed from 10 letters: \(=10*9*8*7\)

Answer: 6 to 1 or E

Number of ways to form 5-letter code: 10!/5! = 10*9*8*7*6
Number of ways to form 4-letter code: 10!/6! = 10*9*8*7

Ratio: 6 to 1

Answer : E

Hi Bunnel,

In this problem you have used Permutations, but in the problem you have used combination, which also deals with code

a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html

Can you please when to use permutaions or Combinations in these type of problems?

In this case the order of the letters matters, but in other question we are only interested in codes which are in alphabetical order (so we are interested in only one particular order).

This post might help: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html#p1150091
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Hello Bunuel,

Thank you very much for responding.

My problem with your initial calculation is that when i do understand how you got 6! in the bottom part of your example above.

When i calculate it, it goes 10 Choose 5 is = 10*9*8*7*6 / 5*4*3*2*1

And then 10 Choose 4 i= 10*9*8*7 / 4*3*2*1

when I calculate 10 Choose 5 / 10 Choose 4.... I get 6/5

I do not know where the error is coming from

Thanks alot

Please use correct notations...

\(P^5_{10}=\frac{10!}{5!}\)

\(P^4_{10}=\frac{10!}{6!}\)

\(\frac{P^5_{10}}{P^4_{10}}=\frac{10!}{5!}*\frac{6!}{10!}=\frac{6}{1}\).

Theory on Combinations: math-combinatorics-87345.html
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Take 1 case: ABCDE for 5 letter code and ABCD for 4 letter code. You Choose 5 at a time and 4 at a time and order is important. So the formula is nPr. Therefore the ratio is 10P5/ 10P4 = 6:1
Hence E.
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Hi All,

To calculate the number of 5-letter codes and 4-letter codes, we have to set up 2 permutations. There's a 'shortcut' though - since the answer choices are RATIOS, we don't actually have to calculate the total number of each type of code.

Total 5-letter codes = (10)(9)(8)(7)(6)
Total 4-letter codes = (10)(9)(8)(7)

Notice how the number of 5-letter codes is the total of 4-letter codes multiplied by 6. Thus, the ratio of codes is 6:1

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi Bunuel,

I struggle with the the order of \(P^5_{10}\). From the theory I understood that the permutation is defined as: \(P^n_{k}\) n being the set of distinct objects and k being the number of objects chosen. Why is this upside down?

Thanks!

\(P^5_{10}\), \(P^{10}_5\), 10P5 are all the same: choosing 5 out of 10, when the order matters. Just different ways of writing the same. Could it be choosing 10 out of 5?
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Permutation problem without repeats.

5-letter word permutations: 10 x 9 x 8 x 7 x 6
4-letter word permutations: 10 x 9 x 8 x 7

5-letter:4-letter = 6:1

Answer is E.
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Although we could use the permutation formula to answer this question, we can also solve using the Fundamental Counting Principle (FCP, aka the slot method). In fact, we can solve any permutation question using the FCP.

Number of 5-letter words we can make
We can select the 1st letter in 10 ways
We can select the 2nd letter in 9 ways
We can select the 3rd letter in 8 ways
We can select the 4th letter in 7 ways
We can select the 5th letter in 6 ways
So the total number of ways to construct a 5-letter word = (10)(9)(8)(7)(6)

Number of 4-letter words we can make
We can select the 1st letter in 10 ways
We can select the 2nd letter in 9 ways
We can select the 3rd letter in 8 ways
We can select the 4th letter in 7 ways
So the total number of ways to construct a 4-letter word = (10)(9)(8)(7)

Ratio of the number of 5-letter code words to the number of 4-letter code words = (10)(9)(8)(7)(6)/(10)(9)(8)(7) = 6/1 = 6 to 1

Answer: E

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

RELATED VIDEO

Of 10 different possibilities.

(I choose 5, different)/(I choose 4, different)

(E1 E2 E3 E4 E5)/(E1 E2 E3 E4)

In the numerator we have 5 spaces and in the denominator 4 spaces.

Each space must be filled with the 10 different letters indicated, keeping in mind that the letters must be different.

In both the numerator and the denominator, E1 can be filled by any of the 10 letters, E2 can be filled by only 9 of the letters (it already occupied a letter in E1, since all the letters must be different) and so on.

(10x9x8x7x6)/(10x9x8x7)
6/1

Answer E

The total number of 5 codes from the 10 letters = 10C5*5!

The total number of 4 letter codes from 10 letters = 10C4*4!

The required ratio = 10C5*5!/10C4*4! = 6

Thus, the correct option is E.

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