Question

     I-Multiplication Rules

1. How many different slats can be made. If the splint is composed of 4 letters and 3 digits.

2. How many special shuttle crews can be formed if: for pilot position, co-pilot and flight engineer there are (8) eight candidates, for two scientists one for solar experiment and one for stellar experiment there are (6) candidates and for two Civilians there are (9) candidates.

II-permutations and combinations

1. In a raffle where there are 10 possible numbers in each ball, if three pellets are extracted. How many ways is it possible to combine extracted numbers?

2. Ten people reach a row at the same time. How many ways can they be formed?

3. In a Olympiad there are 10 swimmers in a race, how many ways can arrive the first three places?

4. How many committees of three teachers can be made if there are 6 teachers to choose from?

Answer 1

Solution

Back-up Theory

If an Activity 1 can be done in n ways, another Activity 2 can be done in m ways and for every one way of doing Activity 1, there are m ways of doing Activity 2, then Activity 1 and Activity 2 can be simultaneously done in (n x m) ways. This is the Rule of Multiplication applicable to both Permutations and Combinations............................. (1)

Number of ways of arranging n distinct things among themselves (i.e., permutations)

= n!

= n(n - 1)(n - 2) …… 3.2.1……………………………………………………..............................................................….….(2)

Values of n!can be directly obtained using Excel Function: Math & Trig FACT (Number)........................................... (2a)

Number of ways of selecting r things out of n things is given by ⁿC_r = (n!)/{(r!)(n - r)!}…...........................................…(3)

Values of ⁿC_r can be directly obtained using Excel Function: Math & Trig COMBIN……............................................ (3a)

Number of ways of selecting r things out of n things when the same thing can be selected any number of times

(i.e., with replacement) is given by n^r........................................................................................................................... (4)

Now, to work out the solution,

NOTE:

All solutions are arrived at assuming no repetition.

Problem I

Part (1)

Vide (3),

4 letters out of 26 letters can be chosen in ²⁶C₄ = 14950 ways [vide (3a)] ................................................................ (5)

3 digits out of 10 digits, 0 included, can be picked in ¹⁰C₃ = 120 ways [vide (3a)] .................................................... (6)

Since for every selection of 4 letters, selection of 3 digits is 120, vide (1), (5) and (6), total number of possible combinations of 4 letters and 3 digits = 14950 x 120 = 1794000 ................................................................................ (7)

Since there is no restriction on the order in which these 4 letters and 3 digits should appear, each of these combinations can be arranged among themselves in 7! = 5040 ways. [vide (2) and (2a)]

Thus, the final answer = 1794000 x 5040 = 9041760000 Answer 1

Part (2).

Out of 8 candidates eligible for selection, 3 flight crew consisting of a pilot, a co-pilot and a flight engineer, can be selected in ⁸C₃ = 280 ways [vide (3) and (3a)] ............................................................................................ (8)

Out of 6 candidates eligible for selection, two scientists can be selected in ⁶C₂ = 15 ways [vide (3)] ..................... (9)

Out of 9 candidates eligible for selection, two civilians can be selected in ⁹C₂ = 36 ways [vide (3)] ...................... (10)

(1), (8), (9) and (10) =>

Number of special shuttle crews that can be formed = 280 x 15 x 36 = 151200   Answer

DONE