Number of permutations when â€˜râ€™ objects are chooses out of â€˜nâ€™ different objects.
$\displaystyle \,{}_nP_r\,$ = $\displaystyle \frac{n!}{(n â€“ r)!}$
1 boy can be selected from 15 boys in _{15}P_{1} ways = $\displaystyle \frac{15!}{(15 â€“ 1)!} = \frac{15 \times 14!}{14!}$ = 15 ways.
1 girl can be selected from 20 girls in _{20}P_{1} ways = $\displaystyle \frac{20!}{(20 â€“ 1)!} = \frac{20 \times 19!}{19!}$ = 20 ways.
Total number of ways by fundamental principle of counting = m x n ways => mn ways => 15 x 20 = 300 ways.
As there is no restriction on repetition, the numbers can be repeated.
There are 5 digits in 1, 2, 3, 4, 5.
We have to arrange 5 digits in 3 places [ here n = 5, r = 3 ]
Then, we can arrange units place digit in 5 ways, tens place digit in 5 ways, hundreds place digit in 5 ways.
Hence total number of required 3 digit number = 5 x 5 x 5 = 125 numbers.
[or] Directly => If repetition is allowed numbers that can be formed = n^{r} => 5^{3} = 125 numbers.
All these numbers have three digits.
5 is in the units place, the middle digit can be any one of the 10 digits, i.e 0 to 9.
Hundredths place digit can be any one of the 9 digits from 1 to 9.
Therefore hundreds place digit can be arranged in 9 ways, tens place digit can be arranged in 10 ways, units place digit can be arranged in 1 way [only with 5].
Required numbers = 9 x 10 x 1 = 90 ways.
$\displaystyle \,{}_nP_r\, = \,{}_5P_n\, = \frac{5!}{(5 â€“ 4)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{1!}$ = 120 ways.
The word "SIGNATURE" have 9 letters and 4 vowels and 5 consonants.
If all the vowels be together, then all vowels are considered as one unit.
S G N T R [A E I U]
5 consonants as 5 units total 5 + 1 = 6 units or objects. These can be arranged in 6! Ways.
And 4 vowels can be arranged among them self in 4! Ways.
Then total number of ways = 6! x 4! = 720 x 24 = 17280 ways.
There are 9 letters in the word "SIGNATURE" with 5 consonants and 4 vowels.
If no two vowels be together, vowels can be placed in 6 places between consonants as shown below.
[Vowel = V, Consonant = C] => V C V C V C V C V C V.
The letters S G N T R can be arranged in 5! Ways, and 4 vowels in _{6}P_{4} ways
Total number of ways = 5! x _{6}P_{4} = 120 x 360 = 43200. [_{6}P_{4} = 360]
Here the retraction is the number must be greater than 4000. Therefore the number in thousands place digit can be filled with 4, 5, 6, 7, 8 or 9. Then the first digit can be chosen in 6 different ways.
The rest of the digits can be chosen from the rest 8 numbers.
It can be in _{8}P_{3} ways = $\displaystyle \frac{8!}{( 8 â€“ 3 )!} = \frac{8 \times 7 \times 6 \times 5!}{5!} $ = 336.
Total number of 4 digit numbers greater than 4000 = 6 x 336 = 2016.
Circular permutation of â€˜nâ€™ objects = (n â€“ 1)!
Out of 8 boys two particular boys may sit together, there are (8 - 1) = 7 objects.
Then circular permutation = (7 â€“ 1)! = 6! Ways = 720 ways.
The two particular boys can be seated in 2! Ways.
Total number of ways 2! x 720 = 1440 ways.
He can go by 8 trains in 8 ways, and can return by 7 trains.
Then total number of ways = 8 x 7 = 56 ways.
As each coin has two faces, then total number of ways = 2^{6} ways = 64 ways.
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Defination: The ways of arranging or selecting smaller or equal number of persons or objects from a group of persons or collection of objects with due regard being paid to the order of arrangement or selection is called Permutation.
1. Fundamental principle of counting: If an event can happen in exactly â€˜mâ€™ ways and if following it, a second event can happen in exactly â€˜nâ€™ ways then the two events in succession can happen in exactly â€˜m x nâ€™ => mn ways.
2. â€˜nâ€™ number of objects can be arranged in n! ways. n! read as â€˜n factorialâ€™.
n! denotes the product of the first â€˜nâ€™ natural numbers.
n! = n x (n â€“ 1) x (n â€“ 2) x (n â€“ 3) x ......
Ex: In how many ways 5 persons can be seated => Answer in 5! ways = 5 x 4 x 3 x 2 x 1 => 120 ways.
3. Number of permutations when â€˜râ€™ objects are chosen out of â€˜nâ€™ different objects.
$\displaystyle \,{}_nP_r\,$ = $\displaystyle \frac{n!}{(n â€“ r)!}$
Ex: How many 3 digit numbers can be formed out of the digits 1, 2, 3, 4, 5 and 6, if no digit is repeated in any number.
n = 6, r = 3, then $\displaystyle \,{}_6P_3\,$ = $\displaystyle \frac{6!}{(6 â€“ 3)!} = \frac{6 \times 5 \times 4 \times 3!}{3!}$ = 120 numbers.
4. Circular Permutation: When we have to arrange â€˜nâ€™ objects along a closed curve or along a circle. The number of circular permutations = ( n â€“ 1 ) ! ways
Ex: How many ways 5 persons can sit around a table?
Sol: ( 5 â€“ 1 )! = 4! = 4 x 3 x 2 x 1 = 24 ways.
5. Number of permutations when some of the things are alike, taken all at a time.
Note: 0! = 1
=> $\displaystyle \frac{n!}{n1!. n2!. n3! â€¦} \ $.
Defination: The number of ways in which smaller or equal number of things are arranged or selected from a collection of things where the order of selection or arrangement is not important.
1. Number of combinations = $\displaystyle \,{}_nC_r\,$ = $\displaystyle \frac{n!}{( n â€“ r )! r!}$
2. Combinations of n different things taking some or all of â€˜nâ€™ things at a time = ( 2^{n} â€“ 1) ways.
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