In a committee of 6 members having equal number of male and female members.
Then the committee consists with 3 male and 3 female members.
3 males can be selected in _{8}C_{3} ways.
3 female can be selected in _{6}C_{3} ways
Required number of ways = _{8}C_{3} x _{6}C_{3} = 56 x 20 = 1120 ways.
The student can chose question From Part A:
_{8}C_{6} = $\displaystyle \frac{8!}{(8-6) \ !6!} = \frac{8 \times 7 \times 6!}{2! \times 6!} = \frac{58}{2}$ = 28 ways.
From Part B:
_{8}C_{6} ways = 28 ways.
Total = 28 x 28 = 784 ways.
If Question number 1 is compulsory the question can be chosen in _{1}C_{1} ways = 1.
Remaining 5 questions out of 9 questions in = _{9}C_{5} ways.
Therefore the student can chose the questions in _{1}C_{1} x _{9}C_{5} = 1 x 126 = 126 ways.
He can dress in _{3}C_{1} x _{5}C_{1} x _{6}C_{1} = 3 x 5 x 6 = 90 ways.
Formula => n â€“ _{1}C_{r} â€“ 1 [or] n â€“ 1 (C_{r â€“ 1}).
8 â€“ _{1}C_{4 â€“ 1} = _{7}C_{3} = $\displaystyle \frac{7 \times 6 \times 5 \times 4!}{4! \ 3 \times 2 \times 1}$ = 35.
Number of shake hands = $\displaystyle \,{}_{10}C_2\, \ = \frac{10!}{(10-2)! \ 2!} = \frac{10 \times 9 \times 8!}{8! \ 2!}$ = 9 x 5 = 45 shake hands.
[OR]
Number of shake hands = $\displaystyle \frac{n (n â€“ 1)}{2} = \frac{10 (10 â€“ 1)}{2} = \frac{90}{2}$ = 45 shake hands.
Required number of ways = $\displaystyle \frac{7!}{4! \ 3!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1}$ = 35 ways.
Number of digits = 6, 3 occupy 2 times, 2 occupy 2 times, 1 occupy 1 time.
Required number of ways = $\displaystyle \frac{6!}{2! \ 2! \ 2!} = \frac{6 \times 5 \times 4 \times 3 \times 2!}{2 \times 2 \times 2}$ = 90 ways.
Hexagon has 6 vertices, no three of them are collinear.
Any two points can be joined to form a straight line.
Number of straight lines which can be formed by joining these 6 points:
=> _{6}C_{2} = $\displaystyle \frac{6 \times 5 \times 4!}{4! \times 2!}$ = 15 straight lines.
Out of 15, 6 are sides = the number of diagonals = 15 â€“ 6 = 9.
A pentagon has 5 vertices, and no three of them are collinear.
Triangles are formed by joining any three of the five vertices of the pentagon.
Required number of triangles = _{5}C_{3} = $\displaystyle \frac{5!}{(5-3)! \ 3!} = \frac{5 \times 4 \times 3!}{2! \ 3!}$ = 5 x 2 = 10.
10 triangles can be formed.
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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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