All 5! and itâ€™s more have 0 as its units place digit.
Sum from 1! to 4! = 33.
Therefore units place digit will be 3.
One specific digit can be chosen in _{1}P_{1} ways, and remaining 3 digits in _{5}P_{3} ways.
Total => _{1}P_{1} x _{5}P_{3} = 1 x 20 = 20 ways.
Number of combinations = $\displaystyle \,{}_nC_r\,$ = $\displaystyle \frac{n!}{( n â€“ r )! \ r!}$
Required number of ways = $\displaystyle \,{}_{16}C_{12}\,$ = $\displaystyle \frac{16!}{(16 â€“ 12)! \ 12!}$
=> $\displaystyle \frac{164 \times 153 \times 147 \times 13 \times 12!}{4 \times 3 \times 2 \times 1 \times 12!}$ = 4 x 5 x 7 x 13 = 1820.
The box contains 4 Red balls and (6 + 5) = 11 non Red balls, if three balls are drawn Possibilities of selecting:
[i] At least 1 Red ball => 1 Red 2 non Red balls => number of ways = _{4}C_{1} x _{11}C_{2} = 4 x 55 = 220 ways.
[ii] 2 Red balls and 1 non Red ball = _{4}C_{2} x _{11}C_{1} = 6 x 11 = 66 ways.
[iii] 3 Red balls = _{4}C_{3} x _{11}C_{0} => $\displaystyle \frac{4!}{(4-3)! \ 3!}$ = 4 x 1 = 4 ways.
Total number of ways = 220 + 66 + 4 = 290.
Red 3; Blue 4; White 6.
One is each color can be chosen in _{3}C_{1} x _{4}C_{1} x _{6}C_{1} = 3 x 4 x 6 = 72 ways.
The student can select 4 subjects out of 8 subjects.
In _{8}C_{4} ways = $\displaystyle \frac{8!}{(8 â€“ 4)! \ 4!} = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4 \times 3 \times 2 \times 1 \times 4!} $ = 2 x 7 x 5 = 70 ways.
First dice is thrown in 6 ways; second dice is thrown in 6 ways.
By multiplication rule = m x n = mn ways
The dice can be thrown in 6 x 6 = 36 ways.
Combinations of n different things taking some or all of â€˜nâ€™ things at a time = (2^{n} â€“ 1) ways.
He has to select one or more of his 6 friends in (2^{6} â€“ 1) = 64 â€“ 1 = 63 ways.
Prime numbers between 60 to 100 => 61, 67, 71, 73, 79, 83, 89, 97 => total 8 numbers.
Total number of pairs = _{8}C_{2} = $\displaystyle \frac{8!}{(8-2)! \ 2!} = \frac{8 \times 7 \times 6!}{6! \ 2!}$ = 28.
3 consonants can be selected out of 5 consonants in _{5}C_{3} ways.
2 vowels can be selected out of 4 vowels in _{4}C_{2} ways.
Hence required number of words, now each selection contains 3 consonants and 2 vowels total 5 letters, 5 letters can be arranged in 5! Ways.
Required number of ways = _{5}C_{3} x _{4}C_{2} x 5! = 10 x 6 x 120 = 7200 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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