Some restricted combinations
1. Number of combinations of n things taken ‘r’ at a time in which x particular things always occur is n-x^Cr-x.
2. Number of combinations of n things taken ‘r’ at a time in which x particular things never occur is n-x^Cr.
3. Number of ways of selections of zero or more things from a group of ‘n’ distinct things in 2^n. i.e nC0+nC1+nC2+…..+nCn=2^n.
4. Number of ways of selection of one or more things from a group of ‘n’ distinct things is 2^n-1. i.e nC1+nC2+…..+nCn=2^n-1 (therefore nC0=0)
5. Number of ways of selection of ‘r’ things out of ‘n’ identical things is 1. (r<=n)
6. Number of ways of selection of zero or more things from a group of ‘n’ identical things is (n+1).
7. Number of ways of selection of some or all things i.e atleast 1 out of (p+q+r+…) things of which p are alike of one kind, q are alike of second kind r are alike of third kind and so on, is {9p+1)(q+1)9r+1)….}-1.
8. Total number of ways of selecting one or more things from ‘p’ identical things of one kind, ‘q’ identical things of second kind, ‘r’ identical things of third kind and n different things is {(p+1)(q+1)(r+1)2^n}-1
9. Number of selections of k consecutive things out of n things in a row is (n-k+1).
10. Number of selection of k consecutive things out of n things in a circle = {n when k<n} & {1 when k=n}.
Problems
1. Given 3 different red dyes. 4 different blue dyes and 5 different green dyes, how many combinations of dyes can be made taking at least one green and one blue dye?
Solution:
Atleast 1 green dye can be selected out of 5 green dyes in (2^5-1) i.e., 31 ways.
Similarly, atleast one blue dye can be selected out of 4 in (2^4-1) ie., 15 ways. And at least 1 red or no red can be selected out of 3 red dyes in 2^3 i.e., 8 ways
Hence the required number of ways = 31x15x8 = 3720
2. In how many ways is it possible to make a selection by taking any number of 15 fruits, namely 3 oranges, 5 apples and 7 mangoes?
Solution:
Out of 15 fruits, 7 are alike of one kind, 5 are alike of a second kind and 3 are alike of a third kind. Hence the required number of ways. = [(7+1)(5+1)(3+1)-1]=191
3. A bag contains 4 mangoes and 5 oranges. In how many ways can I make a selection so as to take atleast one mango and one orange?
Solution:
Atleast one mango cab ne selected in 2^4-1 = 15 ways and atleast one orange can be selected in 2^5-1= 31 ways
Hence the required number of ways = 15x31= 465
4. In how many ways is it possible to make a selection by taking any number of all 20 fruits, namely 9 mangoes, 7 oranges and 4 apples?
Solution:
Required number of ways = [(9+1)(7+1{4+1)]-1 = 399
1. Number of combinations of n things taken ‘r’ at a time in which x particular things always occur is n-x^Cr-x.
2. Number of combinations of n things taken ‘r’ at a time in which x particular things never occur is n-x^Cr.
3. Number of ways of selections of zero or more things from a group of ‘n’ distinct things in 2^n. i.e nC0+nC1+nC2+…..+nCn=2^n.
4. Number of ways of selection of one or more things from a group of ‘n’ distinct things is 2^n-1. i.e nC1+nC2+…..+nCn=2^n-1 (therefore nC0=0)
5. Number of ways of selection of ‘r’ things out of ‘n’ identical things is 1. (r<=n)
6. Number of ways of selection of zero or more things from a group of ‘n’ identical things is (n+1).
7. Number of ways of selection of some or all things i.e atleast 1 out of (p+q+r+…) things of which p are alike of one kind, q are alike of second kind r are alike of third kind and so on, is {9p+1)(q+1)9r+1)….}-1.
8. Total number of ways of selecting one or more things from ‘p’ identical things of one kind, ‘q’ identical things of second kind, ‘r’ identical things of third kind and n different things is {(p+1)(q+1)(r+1)2^n}-1
9. Number of selections of k consecutive things out of n things in a row is (n-k+1).
10. Number of selection of k consecutive things out of n things in a circle = {n when k<n} & {1 when k=n}.
Problems
1. Given 3 different red dyes. 4 different blue dyes and 5 different green dyes, how many combinations of dyes can be made taking at least one green and one blue dye?
Solution:
Atleast 1 green dye can be selected out of 5 green dyes in (2^5-1) i.e., 31 ways.
Similarly, atleast one blue dye can be selected out of 4 in (2^4-1) ie., 15 ways. And at least 1 red or no red can be selected out of 3 red dyes in 2^3 i.e., 8 ways
Hence the required number of ways = 31x15x8 = 3720
2. In how many ways is it possible to make a selection by taking any number of 15 fruits, namely 3 oranges, 5 apples and 7 mangoes?
Solution:
Out of 15 fruits, 7 are alike of one kind, 5 are alike of a second kind and 3 are alike of a third kind. Hence the required number of ways. = [(7+1)(5+1)(3+1)-1]=191
3. A bag contains 4 mangoes and 5 oranges. In how many ways can I make a selection so as to take atleast one mango and one orange?
Solution:
Atleast one mango cab ne selected in 2^4-1 = 15 ways and atleast one orange can be selected in 2^5-1= 31 ways
Hence the required number of ways = 15x31= 465
4. In how many ways is it possible to make a selection by taking any number of all 20 fruits, namely 9 mangoes, 7 oranges and 4 apples?
Solution:
Required number of ways = [(9+1)(7+1{4+1)]-1 = 399
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