# “Forbidden Choices” on the GMAT

by on November 10th, 2012

A business executive is packing for a conference. He has 5 jackets, 4 pairs of shoes, 3 pairs of pants, 2 suitcases and a carry bag. He needs to choose 1 jacket, 1 pair of shoes, and 1 pair of pants to wear on the flight, and one piece of luggage (suitcase or carry bag) to carry the rest of his clothes. How many combinations of clothing and luggage can he choose from?

The solution to this question is fairly simple:

The executive has 5 choices for the jacket AND 4 choices for shoes AND 3 choices for pants AND 2 choices for suitcases or 1 choice for a carry bag.

So, the total number of combinations of clothing and luggage are:

This was a bit too easy. Wasn’t it?

Let’s complicate the problem a bit:

Our fashion-minded executive realizes that one of the pants clashes with two of jackets, and there’s no way he can wear these items together.

How many combinations of clothing and luggage does he have now?

How would you go about calculating this?

How about you calculate the number of “good” choices without the offending articles – first jackets without the pants, then the pants without the jackets, and then add the numbers?

Don’t, seriously just don’t!

There is a much simpler way to find the answer.

What we want to calculate is:

Good combinations  = [Total combinations - "Forbidden" combinations]

Calculate the number of combinations involving “forbidden choices” (the clashing pants AND 2 jackets) as follows:

1) Break down the question {number of forbidden choices} into a series of boxes, assign one box to each “item”.

Forbidden choices:
Jacket    Shoes    Pants    Suitcase    Carry Bag

2) For each box, find the number of choices available for that item.

Remember that we’re calculating the number of “forbidden choices”, so the number of choices for the pants and jackets will include ONLY the forbidden articles (2 J, 1 P):

Forbidden choices:
Jacket: 2  Shoes: 4    Pants: 1    Suitcase: 2    Carry Bag: 1

3) Multiply/add the number of choices from each source according to its relationship (AND/OR):

Forbidden choices:

Jacket AND Shoes AND Pants AND Suitcase OR Carry Bag
= 2 x 4 x 1 x (2 + 1) = 24

Remember that these are 24 “forbidden combinations” – combinations containing the clashing pants AND the clashing jackets.

To find out how many “good combinations” we have, subtract 24 from the total of 180 to get:

[Total combinations - "forbidden" combinations]
= 180 – 24
= 156 combinations.

## Remember:

When a problem introduces forbidden choices, the calculation method is as follows:

[Total combinations - "Forbidden" combinations] = Good combinations

1) Calculate the total # of combinations using the regular Step-By-Step method.

2) Calculate the total # of forbidden combinations using the same method. Remember to include the other sources (the ones without a forbidden choice in them) as well.

3) Subtract the # of forbidden combinations from the total # of combinations to find the # of good combinations.

Now, use the above method to solve the following question:

How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?

20
150
225
300
320

• 225 C

• Hi,

Its 300 D

Rgds,
Amit

• I also got 225, C.

• c.225

• C. 225

Here goes the explanation:

1st digit choices: Even 2,4,6,8 = 4

2nd digit choices: Odd 1,3,5,7,9 =5

3rd digit choices: Prime 2,3,5,7 (remember 1 is not prime) =4

4th digit choices: Divisible by 3: 3,6,9 =3

Total no. of ways = 4 x 5 x 4 x 3 = 240

But 2 cannot be used more than once

therefore, forbidden ways = 1 x 5 x 1 x 3 = 15

good ways = 240 - 15 = 225

Another way of solving:

No. of ways = 4 x 5 x 3 x 3 = 180 (we used '2' here in 1st digit so dint use in 3rd digit)

additionally, when we use '2' in 3rd digit, no. of ways = 3 x 5 x 1 x 3 = 45 (3 ways in 1st digit since we are not using '2' there)

total ways = 180 + 45 = 225

• I believe the answer is D. 300

Choices
1st digit : 2,4,6,8 = 4
2nd digit : 1,3,5,7,9 = 5
3rd digit : 2,3,5,7 = 4
4th digit : 0,3,6,9 = 4 (Don't forget zero. It's divisible by 3)

No.of ways : 4 x 5 x 4 x 4 = 320

Forbidden:
"2 cannot be used more than once"

Thus, the forbidden choices = 1 x 5 x 1 x 4 = 20

Required combinations = 320 - 20 = 200

• ok ok ya 0 should also be considered. thanks!

• I also got D - 300.

For the 4th digit choices (numbers that are divisible by 3), I also included 0.

POSSIBLE CHOICES

1st Digit (Even): 2,4,6,8 (4 possibilities)
2nd Digit (Odd): 1,3,5,7,9 (5 possibilities)
3rd Digit (Prime): 2,3,5,7 (4 possibilities)
4th Digit (Divisible by 3): 0,3,6,9 (4 possibilities)

For 4 x 5 x 4 x 4 =320 possibilities

"FORBIDDEN" CHOICES

1st Digit (Even): 2 (1 possibility)
2nd Digit (Odd): 1,3,5,7,9 (5 possibilities)
3rd Digit (Prime): 2 (1 possibility)
4th Digit (Divisible by 3): 0,3,6,9 (4 possibilities)

For 1 x 5 x 1 x 4 = 20 choices

"GOOD" CHOICES = 320 - 20 = 300 choices.

• Isn't zero an even number, as well? If so, the first digit would have five possiblities, rendering no correct answer to choose from. 1) 5 even, 2) 5 odd, 3) 4 prime, and 4) 4 divisible by 3. This would equal 400. The number of forbidden choices would then be 1x5x4x1=20. 400-20=380.

• I didn't include 0 as a possibility in the even number because if 0 was the first digit, it would be a 3-digit number and not a 4-digit number.

I.e. 0320 = 320 = 3-digit number. When they say a "4-digit" number, my interpretation is that the first digit CANNOT be 0.

• haha... True story. Man, that's embarrassing. No more jumping on beat the gmat while at work.

• 300 (D)

Total combinations: 4 (even: 2,4,6,8) * 5 (odd: 1,3,5,7,9) * 4 (prime: 2,3,5,7) * 4 (divisible by 3: 0,3,6,9): 320

Subtract the Combinations that break the requirements: 1 (2 is picked) * 5 (any odd) * 1 (another 2 is picked) * 4 (any divisible by 3): 20

320-20= 300