 Mathematics PERMUTATIONS AND COMBINATIONS - Fundamental Principle of Counting
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### Fundamental Principle of Counting color(red)=>"Let us consider the following problem. "

color(red)"Problem. 1 ." Mohan has color(blue)("3 pants") and color(blue)("2 shirts.") How many color(blue)("different pairs") of a pant and a shirt, can he dress up with ?

There are color(blue)("3 ways in which a pant can be chosen"), because there are 3 pants available.

Similarly, color(blue)("a shirt can be chosen in 2 ways"). For every choice of a pant, there are 2 choices of a shirt.

Therefore, there are color(red)(3 × 2 = 6) pairs of a pant and a shirt.

Let us name the three pants as P_1, P_2 , P_3 and the two shirts as S_1, S_2. Then, these six possibilities can be illustrated in the Fig. 1.

color(red)(=>)"Let us consider another problem of the same type."

color(red)"Problem. 2 ." Sabnam has color(blue)("2 school bags"), color(blue)("3 tiffin boxes") and color(blue)("2 water bottles"). In how many ways can she carry these items ( choosing one each ).

color(blue)("A school bag can be chosen in 2 different ways.")

After a school bag is chosen, color(blue)("a tiffin box can be chosen in 3 different ways.")

Hence, there are 2 × 3 = 6 pairs of school bag and a tiffin box.

For each of these pairs color(blue)("a water bottle can be chosen in 2 different ways.")

Hence, there are color(red)(6 × 2 = 12) different ways in which, Sabnam can carry these items to school.

If we name the 2 school bags as B_1, B_2, the three tiffin boxes as T_1, T_2, T_3 and the two water bottles as W_1, W_2, these possibilities can be illustrated in the Fig. 2.

color(blue)(★ ul" Fundamental Principle of Counting")

“ If an event can occur in color(blue)(m) different ways, following which another event can occur in color(blue)(n) different ways, then the total number of occurrence of the events in the given order is color(red)(m×n).”

color(red)("For any finite number of events. For example, for 3 events, the principle is as follows:")

=> ‘If an event can occur in color(blue)(m) different ways, following which another event can occur in color(blue)(n) different ways, following which a third event can occur in color(blue)(p) different ways, then the total number of occurrence to ‘the events in the given order is color(blue)("m × n × p.”)

color(red)("In the first problem"), the required number of ways of wearing a pant and a shirt was the number of different ways of the occurrence of the following events in succession:
(i) the event of choosing a pant

(ii) the event of choosing a shirt.

color(red)("In the second problem"), the required number of ways was the number of different ways of the occurrence of the following events in succession:
(i) the event of choosing a school bag

(ii) the event of choosing a tiffin box

(iii) the event of choosing a water bottle.

Here, in both the cases, the events in each problem could occur in various possible orders. But, we have to choose any one of the possible orders and count the number of different ways of the occurrence of the events in this chosen order.
Q 3068823705 Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed. Solution:

There are as many words as there are ways of filling in 4 vacant places square , square , square , square by the 4 letters, keeping in mind that the repetition is not allowed. The first place can be filled in 4 different ways by anyone of the 4 letters R,O,S,E. Following which, the second place can be filled in by anyone of the remaining 3 letters in 3 different ways, following which the third place can be filled in 2 different ways; following which, the fourth place can be filled in 1 way. Thus, the number of ways in which the 4 places can be filled, by the multiplication principle, is 4 × 3 × 2 × 1 = 24. Hence, the required number of words is 24.

color{red} "Key Point" If the repetition of the letters was allowed, how many words can be formed? One can easily understand that each of the 4 vacant places can be filled in succession in 4 different ways. Hence, the required number of words = 4 × 4 × 4 × 4 = 256.

Q 3088023807 Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other? Solution:

There will be as many signals as there are ways of filling in 2 vacant places in succession by the 4 flags of different colours. The upper vacant place can be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant place can be filled in 3 different ways by anyone of the remaining 3 different flags. Hence, by the multiplication principle, the required number of signals = 4 × 3 = 12.
Q 3068123905 How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated? Solution:

There will be as many ways as there are ways of filling 2 vacant places square square in succession by the five given digits. Here, in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10.
Q 3018234100 Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available. Solution:

A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us count the possible number of signals consisting of 2 flags, 3 flags, 4 flags and 5 flags separately and then add the respective numbers. There will be as many 2 flag signals as there are ways of filling in 2 vacant places in succession by the 5 flags available. By Multiplication rule, the number of ways is 5 × 4 = 20. Similarly, there will be as many 3 flag signals as there are ways of filling in 3 vacant places

The number of ways is 5 × 4 × 3 = 60.
Continuing the same way, we find that
The number of 4 flag signals = 5 × 4 × 3 × 2 = 120
and the number of 5 flag signals = 5 × 4 × 3 × 2 × 1 = 120
Therefore, the required no of signals = 20 + 60 + 120 + 120 = 320. 