MO
MO-ARML Practice
Mathleague / ARML
Team Test: Counting Procedures
Name
Date
______________________________
_______________
1} In how many ways can the eleven letters in ‘MATH CONTEST’ be arranged into eleven-letter ‘words’?
2} If a chain will be strung with 4 red and 4 white beads, how many distinct necklaces can be made from
them?
3} 10 people have been chosen to sit in a row with 10 chairs, and 2 of the people are Guy and Doll. How
many orders of seating are possible if Guy and Doll must sit with at least 2 chairs between them?
4} A computer password consisting of two letters and three digits is to be created. They can be placed in
any order, but no duplications are allowed. For instance, X78P2 and 504YB are both OK. How many
different passwords are possible?
5} Find the number of 4-digit odd numbers with distinctly different digits.
6} The letters D, E, I, L and V can be arranged into 120 different 5-letter ‘words’. If those 120 ‘words’ are
listed alphabetically from positions 1 to 120, then what would be the sum of the positions for the words
‘DEVIL’ and ‘LIVED’?
7} Find the number of ways to distribute three red, four blue, and five green balls into four distinct jars.
8} If 2n is the product of all of the powers of 2 with positive integer exponents which are factors of 200!,
then what is the value of n?
9} Find the number of ways to distribute five different-colored balls into eight boxes if at most one ball can
go into each box.
10} ‘Make the arrangements’ to find the coefficient k in the term kx3y2z5 from the expansion of (x + y + z)10.
Solutions to Team Test: Counting Procedures Problems
1} Since only the letter T repeats, it is a permutation of 11! / 3! = 6 652 800.
6 652 800
2} [Keeping in mind that a chain of beads can be viewed from both sides] If we choose to focus just on the
4 red beads, then the cases are: all separated, which will occur in only 1 way; all together, which will also
occur in only 1 way; separated into two pairs, which can only occur 2 ways — when the whites are in 1-and-3
or 2-and-2 bead arrangements; two are together but the other 2 are separated, which can occur in 2 ways; or 3
are together and the other alone, which can occur in 2 ways. Hence, we have 1 + 1 + 2 + 2 + 2 = 8.
8
3} There are 10! ways to arrange the people. But of those, there are also 9 pairs of seats where G and D
would be next to each other, with 2! ways they could sit in each pair, and 8! ways the rest of the people could
then be seated. Similarly, there are 8 pairings with one chair between G and D. So, the number of seating
arrangements allowed is: 10! – (9 + 8)2! · 8! = (90 – 34) · 8! = 2 257 920.
2 257 920
4} There are 26 C 2 = 325 combinations of letters and 10 C 3 = 120 combinations of digits possible. These
can be arranged into any of 5 positions, so the number of potential passwords is 325 · 120 · 5! = 4 680 000.
4 680 000
5} The number choices can be modeled as ? ? ? 5, which then can be separated into 3 cases: there
might be no zero digits —› 8 · 7 · 6 · 5, or the second digit is zero —› 8 · 1 · 7 · 5, or the third digit is zero —›
8 · 7 · 1 · 5. This yields 1680 + 280 + 280 = 2 240.
2 240
6} Of the 5! different ‘words’ formed, each of the 5 letters is at the beginning of 4! = 24 words, each of the
4 remaining letters is in the second spot of 3! of those, ... . So, ‘DEVIL’ is in position 1 + 0·24 + 0·6 + 2·2 +
0·1 = 5, and ‘LIVED’ is in position 1 + 3·24 + 2·6 + 2·2 + 1·1 = 90. The sum is 5 + 90 = 95.
95
7}
6C3
·
7C3
·
8C3
= 20· 35· 56 = 39 200.
39 200
8} Of the integers from 1 to 200 which multiply to make the product 200!, the 100 evens contain at least
one factor of 2; 50 multiples of 4 contain a second factor of 2; 25, a third; 12, a 4th; etc. This produces a sum
of 100 + 50 + 25 + 12 + 6 + 3 + 1 = 197, so the largest factor of 200! is 2197. For 2n to be the product of all of
the factors from 21 to 2197, n = 197 + (196)(197)/2 = 19 503.
19 503
9} Three boxes will go empty in 8 C 3 ways. Since the balls are distinguishable, multiply that by 5!. This
is the same as 8 P 5. So, 8 · 7 · 6 · 5 · 4 = 6 720.
6 720
10} This is equivalent to asking how many ways can we arrange the multiplication of 3 x’s, 2 y’s and 5 z’s.
This is a permutation: k = 10! / (3! 2! 5!) = 2 520.
2 520