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A random 10-letter code is to be formed using the letters...

This topic has 1 expert reply and 2 member replies

A random 10-letter code is to be formed using the letters...

Post Thu Dec 07, 2017 11:02 am
A random 10-letter code is to be formed using the letters A, B, C, D, E, F, G, H, I and I (only the "I" will be used twice). What is the probability that a code that has the two I's adjacent to one another will be formed?

(A) 1/10
(B) 1/8
(C) 1/5
(D) 1/4
(E) 1/2

The OA is C.

I'm confused with this DS question. Please, can any expert assist me with it? Thanks in advanced.

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Post Thu Dec 07, 2017 4:41 pm
A simple way:

Start by treating the two I's as one unit grafted together. If we then arrange the "nine" letters (A, B, C, D, E, F, G, H, and II), we've got 9! arrangements, so there are 9! ways to put the I's together.

But if we have no restrictions at all, we have 10!/2! ways of arranging 10 letters, exactly two of which are identical. (If this formula is new, read up on permutations with repeating elements. The basics are pretty straightforward, and the visuals should help!)

So our answer = I's together / any arrangement = 9! / (10!/2!) = 1/5

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Post Thu Dec 07, 2017 12:32 pm
LUANDATO wrote:
A random 10-letter code is to be formed using the letters A, B, C, D, E, F, G, H, I and I (only the "I" will be used twice). What is the probability that a code that has the two I's adjacent to one another will be formed?

(A) 1/10
(B) 1/8
(C) 1/5
(D) 1/4
(E) 1/2

The OA is C.

I'm confused with this DS question. Please, can any expert assist me with it? Thanks in advanced.
Another approach is to consider successful combinations divided by total combinations.

Number of ways the 10 letters can be arranged starts with 10!. However, because two of the letters, I, are the same, this needs to be divided by 2!.

So total available combinations is 10!/2!.

Now, the pairs of I's can occupy 9 positions. The other 8 letters can be arranged 8! ways.

So, the total successful combinations is 9*8!

Divide the successful combinations by total: 9*8!/(10!/2!) = 9*8!*2!/10! = 9*2!/(10*9)=2!/10 = 1/5, C

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regor60 Master | Next Rank: 500 Posts Default Avatar
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Post Thu Dec 07, 2017 12:25 pm
LUANDATO wrote:
A random 10-letter code is to be formed using the letters A, B, C, D, E, F, G, H, I and I (only the "I" will be used twice). What is the probability that a code that has the two I's adjacent to one another will be formed?

(A) 1/10
(B) 1/8
(C) 1/5
(D) 1/4
(E) 1/2

The OA is C.

I'm confused with this DS question. Please, can any expert assist me with it? Thanks in advanced.
A probability approach is as follows:

Probability of picking an I for the first position: 2/10

Probability of picking the second I in the second spot: 1/9

Therefore, the probability of having an I in the first and second positions: 2/10*1/9

Now, the two I's could be in the second and third positions and so on up to ninth and tenth

Second and third positions would be 8/10 for one of the other letters in the first position, 2/9 for the first I and 1/8 for the second I. Notice that those multiplied together is the same 2/10*1/9 from before. Test this on the remaining positions and the same result occurs.

In all, there are 9 instances of 2/10*1/9 so the total probability is: 9*2/10*1/9 or 1/5, C

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