### CAT Online Coaching - Permutation and Combination, Fixing the Errors II

This post had given a series of questions with incorrect solutions. Given below are
the "debugged" solutions to questions 4, 5 and 6. The questions and the incorrect solutions here

4. In how
many ways can we select 2 cards from a card pack such that both belong to the
same suit?

**Given solution**

The first
card can be selected in 52 ways, as it can be any card in the pack. Now, there
are 12 different ways of selecting the second card. Depending on the suit of
the first card, the choice set we have for the second card gets limited. If the
first card were a club, then the second card must be from among the 12
remaining clubs and so on.

So, total
number of outcomes = 52 * 12.

**Bug in the solution:**

We are
double-counting when we do this. If we select 2H as the first card and 5D as
the second card; this is same as selecting 5D as the first card and 2H as the
second card. However, in our current approach, we will end up counting both of
these.

**Correct solution:**

The
correct approach would be to select one of the four suits first and then select
2 cards from 13 in that suit. Or, the number of ways of doing this = 4 *

^{13}C_{2}= 52 * 6.
5. When a
coin is tossed 10 times, what is the probability of getting more heads than
tails?

**Given solution**

When a
coin is tossed 'n' times, the probability of getting more heads than tails =
Probability of getting more tails than heads. So, we do not need to really
compute this. Probability = 1/2.

**Bug in the solution:**

We are
not counting the scenarios where there are an equal number of heads and tails.
If we threw the coin an odd number of times, the probability of getting more
heads than tails will be ½.

**Correct solution:**

Number of
heads could be 6, 7, 8, 9, or 10. Probability = (

^{10}C_{6}+^{10}C_{7}+^{10}C_{8}+^{10}C_{9}+^{10}C_{10})/ 2^{10}
The other
way of thinking about this would be ½ (1 –

^{10}C_{5})/ 2^{10}. If we removed the scenario where the number of heads and tails are equal, the probability of more heads than tails should be equal to probability of more tails than heads.
6. In how
many ways can we arrange 3 boys and 3 girls on a straight line such that no two
boys stand next to each other

**Given solution**

We can
arrange the 6 people as BGBGBG or GBGBGB. If we arranged them as BGBGBG, we
would have 3! * 3! ways of arranging the 3 boys and the 3 girls. So, total
number of possibilities = 2 * 3! * 3! = 72 ways.

**Bug in the solution:**

We are
not accounting for arrangement such as BGBGGB and BGGBGB. Note that the
question says that no two boys should stand next to each other. Two girls can
stand next to each other.

**Correct solution:**

There are 4
possible outlines – BGBGBG, GBGBGB, BGBGGB or BGGBGB. So, total number of
possibilities = 3! * 3! * 4 = 144.

Labels: CAT 2015, CAT Coaching Online, CAT Combinatorics, CAT Online Preparation, CAT Permutation Combination

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