# GRE® Permutation : Select One or More Answer Choices

Concept: Compute the number of outcomes for each of the mentioned events

### Directions: Select one or more answer choices according to the specific question directions.

If the question does not specify how many answer choices to select, select all that apply.

• The correct answer may be just one of the choices or as many as all of the choices, depending on the question.
• No credit is given unless you select all of the correct choices and no others.

If the question specifies how many answer choices to select, select exactly that number of choices.

#### Question: For which of the following events will the number of outcomes exceed 50?

1. The number of outcomes in which at least three heads appears in 6 consecutive tosses of a fair coin.
2. The number of outcomes in which the sum of the digits that appear on the facing side is odd when a fair die rolled thrice.
3. The number of outcomes in which the two cards drawn from a pack of well shuffled cards are both red and face cards.
4. The number of outcomes in which the vowels appear together when the letters of the word 'PRIORITY' are reordered.
5. The number of ways of posting 6 different letters in 2 different post boxes such that at least one letter is posted in each of the boxes.
6. The number of ways of selecting at least one Indian and at least one American for a debate from a group comprising 3 Indians and 4 Americans and no one else.

Video explanation will be added soon.

#### Choice A

##### The number of outcomes in which at least three heads appears in 6 consecutive tosses of a fair coin.

A coin tossed 6 times consecutively could result in outcomes ranging from 0 heads to 6 heads.
We have to count the outcomes in which at least three heads appear

What does at least three heads mean?

1. Number of outcomes with 3 heads: 6 choose 3 or 6C3 = $$frac{6!}{3! \times 3!}$ = 20. 2. Number of outcomes with 4 heads: 6 choose 4 or 6C4 = $\frac{6!}{4! \times 2!}$ = 15. 3. Number of outcomes with 5 heads: 6 choose 5 or 6C5 = $\frac{6!}{5! \times 1!}$ = 6. 4. Number of outcomes with 6 heads: 6 choose 6 or 6C6 = 1 way. So, number of outcomes in which at least 3 heads will appear = 20 + 15 + 6 + 1 = 42 Choice A is NOT one of the answers. #### Choice B ##### The number of outcomes in which the sum of the digits that appear on the facing side is odd when a fair die rolled thrice. The total number of outcomes when a fair die is rolled thrice is 6 * 6 * 6 = 216. If you list down a few of the outcomes and observe the pattern, you will realize that half the outcomes have a sum that is odd and the other half has a sum that is even. Here is a sample of the first 6. {1, 1, 1, odd}, {1, 1, 2, even}, {1, 1, 3, odd}, {1, 1, 4, even}, {1, 1, 5, odd}, {1, 1, 6, even}. Therefore, ½ of 216 = 108 outcomes have an odd sum. Choice B is one of the answers where the number of outcomes is more than 50. #### Choice C ##### The number of outcomes in which the two cards drawn from a pack of well shuffled cards are both red and face cards. A pack of well shuffled cards has 52 cards. The break up of the pack on different parameters is listed in the table given below Color Suit Numbered cards Face cards Aces Red - 26 cards Diamond - 13 cards$2 to 10) 9 cards (J, Q, and K) 3 cards 1 card
Hearts - 13 cards (2 to 10) 9 cards (J, Q, and K) 3 cards 1 card
Black - 26 cards Spades - 13 cards (2 to 10) 9 cards (J, Q, and K) 3 cards 1 card
Clubs - 13 cards (2 to 10) 9 cards (J, Q, and K) 3 cards 1 card
Total 52 cards 36 numbered cards 12 face cards 4 aces

There are 3 face cards that belong to the diamonds and 3 face cards that belong to hearts that are red and are face cards.
So, if the two cards that are selected are red and face cards, those two should have been selected from this set of 6 cards.
Number of ways of selecting 2 cards out of 6 = 6 choose 2 = 6C2 = $$frac{6!}{4!*2!}$ = 15. Choice C is NOT one of the answers. #### Choice D ##### The number of outcomes in which the vowels appear together when the letters of the word 'PRIORITY' are reordered. PRIORITY is an 8-letter word. The vowels in the word are I, O, and I. The consonants in the word are P, R, R, T, and Y. The condition to be satisfied is that the vowels appear together. So, if we consider the 3 vowels as one unit, the vowels will appear together. Let us call this unit comprising 3 vowels as δ. So, we need to reorder P, R, R, T, Y and δ i.e., 6 letters have to be reordered - out of which two letters are 'R's. So, these can be reordered in $\frac{6!}{2!}$ = 360 ways. Now the unit δ comprising 3 letters I, O, and I can also be reordered. These 3 letters of which two are alike can be reordered in $\frac{3!}{2!}$ = 3 ways. So, the total number of ways to reorder the letters of the word PRIORITY such that the vowels appear together = 360 * 3 = 1080 ways. # Do not waste time computing the final answer In the GRE test, if you have found that the number of ways or reordering P, R, R, T, Y and δ is 360, you should not even bother to spend time computing the number of ways of reordering the unit comprising the vowels because the final answer will be greater than 360 - which is definitely greater than 50. D is one of the choices in which the number of outcomes exceeds 50. #### Choice E ##### The number of ways of posting 6 different letters in 2 different post boxes such that at least one letter is posted in each of the boxes. Let us says the letters are addressed to A, B, C, D, E, and F and the boxes are 1 and 2. Letter A can be posted in the boxes in 2 ways. Either into box 1 or into box 2. Similarly, letter B and the remaining letters can be posted into a box in 2 ways each. Letter A can be posted in 2 ways AND letter B can be posted in 2 ways and so on. So, the 6 letters can be posted in 2 * 2 * 2 * 2 * 2 * 2 = 26 = 64 ways. The only problem is that these 64 ways include the possibility that all the 6 letters are posted into only one of the post boxes - either all 6 into box 1 or all 6 into box 2. If we count these two possibilities out, then we have 64 - 2 = 62 ways in which 6 different letters can be posted into 2 different post boxes such that at least one letter is posted in each of the boxes. Choice E is one of the choices where the number of outcomes exceeds 50. #### Choice F ##### The number of ways of selecting at least one Indian and at least one American for a debate from a group comprising 3 Indians and 4 Americans and no one else. There are 3 Indians in the group. We have to select at least 1 Indian. i.e., we could select 1 or 2 or all 3 Indians. Similarly, there are 4 Americans in the group. We have to select at least 1 American. i.e., we could select 1 or 2 or 3 or all 4 Americans. Number of ways of selecting 1 Indian = 3 choose 1 = 3C1 = 3 ways. Number of ways of selecting 2 Indians = 3 choose 2 = 3C2 = 3 ways. Number of ways of selecting all 3 Indians = 3 choose 3 = 3C3 = 1 way. Hence, the number of ways of selecting one or more Indians =$3 + 3 + 1) = 7 ways.

Number of ways of selecting 1 American = 4 choose 1 = 4C1 = 4 ways.
Number of ways of selecting 2 Americans = 4 choose 2 = 4C2 = $$frac{4!}{2! * 2!}$ = 6 ways. Number of ways of selecting 3 Americans = 4 choose 3 = 4C3 = 4C1 = 4 ways. Number of ways of selecting all 4 Americans = 4 choose 4 = 4C4 = 1 way Hence, the number of ways of selecting one or more Americans =$4 + 6 + 4 + 1) = 15 ways.

So, the number of ways of selecting at least one Indian and at least one American = 7 * 15 = 105 ways.

#### Alternative way to compute "at least one" selection

Let the 3 Indians be A, B, and C. A can either be included in the group or excluded from the group. A has 2 possibilities. So, has B and so has C. Therefore, the 3 Indians may all be selected or none be selected in 2 * 2 * 2 = 23 ways.
However, we need to select at least one Indian. so, the possibility that none of the Indians are selected should be counted out.
Hence, there are (23 - 1) ways

In general, at least one object out of n distinct objects can be selected in (2n - 1) ways.
So, one or more Indians may be selected in (23 - 1) ways = 7 ways and one or more Americans may be selected in (24 - 1) = 15 ways.

Hence, totally there will be 7 * 15 = 105 ways.

Choice F is one of the choices in which the total outcomes exceeds 50.