1)Alan has 5 pairs of trousers and 6 shirts. Ben has 7 pairs of trousers and 4
shirts. Who has the most choice of what to wear? Why?
Alan has 5x6 or 30 outfits.
Ben has 7x4 = 28 outfits.
Alan has more.
2)From the digits 0, 1, 2, 3, 4, 5, how many 4-digit even numbers with distinct
digits can be formed? (Hint: Consider two cases: when it begins with 1, 3 or 5
and when it begins with 2 or 4)
Case 1: It begins with 1,3, or 5.
There are 3 choices for the 1st digit. {1,3,5}
There are 3 choices for the 4th digit. {0,2,4}
There are 4 choices left for the 2nd digit.
There are 3 choices left for the 3rd digit.
3x3x4x3 = 108 ways for Case 1.
Case 2: It begins with a 2 or 4.
There are 2 choices for the 1st digit. {2,4}
There are 2 choices for the 4th digit. (0 and whichever of {2,4} remains
There are 4 choices left for the 2nd digit.
There are 3 choices left for the 3rd digit.
2x2x4x3 = 48 ways for Case 2.
Total: 108+48 = 156 ways
3)Morse code is made from dots and dashes. The letter E is a dot (using 1
character), the letter N is a dash followed by a dot (using 2 characters), the
letter S is three dots (using 3 characters). Explain why it is possible to
construct the 26 letters of the alphabet using maximum 4 characters.
There are 2 ways to use 1 character.
There are 2x2 = 4 ways to use 2 characters.
There are 2x2x2 = 8 ways to use 3 cghartacters
There are 2x2x2x2 = 16 ways to use 4 characters
That's 2+4+8+16 = 30 ways, and there are only 26 letters.
4)5 cards are selected randomly from a packet of 52 cards (there are 4 suits in
the packet, and each suit has 13 cards). In how many ways can at least 4 cards
of the same suit be selected?
Case 1. Exactly 4 of the same suit, and one card of another suit.
Pick the suit 4 ways.
Pick them from the 13 of that suit 13C4 = 715 ways
Pick the odd card from the other 3 suits, which is from 3x13 = 39 cards.
That's 4x715x39 = 111540 ways
Case 2. All 5 of the same suit.
Pick the suit 4 ways.
Pick the 5 cards from the 13 of that suit 13C5 = 1287 ways.
Total for both cases: 111540+1287 = 112827
5)3 girls and 4 boys sit in a row for a photograph. Find the number of ways they
can be seated if
(a) the 3 girls sit together.
Arrange the girls together in a trio in 3! = 6 ways
For each of those ways, we only have 5 "things" to permute, 4 boys and 1
girl-trio, which we can arrange in 5!=120 ways.
Answer: 6x120 = 720 ways.
[You can also look at it that we have 5 types of arrangements:
BBBBGGG,BBBGGGB,BBGGGBB,BGGGBBB,GGGBBBB
and the boys can be arranged 4! ways and the girls 3! ways, 5x24x6 = 720 ways]
(b) no girls sit together.
Put spaces before, after, and between the boys, like this.
__B__B__B__B__
We can then put the 3 girls in any of those 5 spaces.
We can arrange the 4 boys in 4! or 24 ways.
We can pick spaces from the 3 girls in 5C3 = 10 ways.
We can arrange the girls in those 3 spaces in 3! = 6 ways.
Answer: 24x10x6 = 1440 ways
(c) only 2 of the girls sit together.
Choose the girl to sit on the left in the girl-pair in 3 ways
Choose the girl to sit on the right in the girl-pair in 2 ways
As in the (b) part, put spaces before, after, and between the boys,
We can put the single girl and the girl-pair in any of those 5 spaces.
__B__B__B__B__
We can arrange the boys in 4! = 24 ways
We can pick a space for the single girl in 5 ways
We can pick a space for the girl-pair in 4 ways
That's 3x2x24x5x4 = 2880 ways.
(d) the 3 girls sit in alphabetical order, from left to right, but not
necessarily together.
We first calculate the number of ways the 7 people can sit in a row,
which is 7! = 5040. But this is 3! = 6 times too many, because the 5040
counts all the 3!=6 permutations of the girls, and we only want to count
the one permutation when they are in alphabetical order. So we divide
5040 by 6 and get 840 ways.
(e) girls and boys alternate, where the tallest girl sits in the middle.
Let the tallest girl be T, then we have this gender-wise configuration
B,G,B,T,B,G,B
we choose the tallest girl 1 way
We can arrange the boys 4! = 24 ways.
We can arrange the 2 shorter girls 2! = 2 ways.
Answer 1x24x2 = 48 ways
Edwin
