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
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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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10P5 /10P4 .

Why we used Permutations not combinations as here sequence is important (Code)

OA: E

Number of \(5\) letter code words \(= 10*9*8*7*6\)
Number of \(4\) letter code words \(= 10*9*8*7\)

the ratio of the number of \(5\)-letter code words to the number of \(4\)-letter code words \(=\frac{ 10*9*8*7*6}{10*9*8*7}=\frac{6}{1}\)

The ratio of the number of 5-letter code words to the number of 4-letter code words chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J is given by 10P5/10P4 = 6:1

Option E is the correct Ans!!
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5 letter = 10*9*8*7*6
4 letter = 10*9*8*7
ratio ; 6 : 1
IMO E

so ,it goes like this.
for those who got confused.
as, you can choose 10 choose 5 in 10c5 ways and also you have to multiple by 5! as you can shuffle among themselves in 5! ways , do mind order matters here coz you are picking for a code word.
so fianlly 10c5 * 5!/ 10c4 * 4! = 6/1

ORDER MATTERS AND WITH NO OF WAYS OF REORDERING IS 5! .

i hope its clear..