permutation and combination

Question

no of 3 digit odd numbers that can be formed without repetition of digits

Answer 1

,5*(9*8)=360; there are 5 odd digit numbers, so once last digit is chosen (making number odd), you can permutate down from the remaining 9 numbers. Thus there is 8*9 = 72 odd non repeating numbers for each odd digit for three digit numbers. And you can do this five times. 

 

-I'm available for online tutoring. 

Answer 2

To make the number odd, we have one of {0,1,2,3,4,5,6,7,8,9} followed by one of {0,1,2,3,4,5,6,7,8,9} followed by one of {1,3,5,7,9}.  Now, if a leading 0 is not allowed, then it should be removed from the first set.  However, my briefcase lock has 3 dials, so I'll allow 0 (you should check; the answer could be 360-40=320).

 

Now, you disallow repetition of digits.  You have formulas for doing that (and other tutors might use them).  Still keep the two groups (two of all digits) followed by one group (one more digit); this will determine allowed combinations.  See if you can determine the pattern in the list below.

 

For fun, I wrote the following computer program:

  N = 0

  For I = 0 to 9

    For J = 0 to 9

      For K = 1 to 9 Step 2

        If (I<>J) AND (I<>K) AND (I<>K) then N = N + 1; Output I,J,K

      NEXT K

    NEXT J

  NEXT I

Output N

 

The value the computer found was 360 (see list below).

 

1    0 1 3
2    0 1 5
3    0 1 7
4    0 1 9
5    0 2 1
6    0 2 3
7    0 2 5
8    0 2 7
9    0 2 9
10    0 3 1
11    0 3 5
12    0 3 7
13    0 3 9
14    0 4 1
15    0 4 3
16    0 4 5
17    0 4 7
18    0 4 9
19    0 5 1
20    0 5 3
21    0 5 7
22    0 5 9
23    0 6 1
24    0 6 3
25    0 6 5
26    0 6 7
27    0 6 9
28    0 7 1
29    0 7 3
30    0 7 5
31    0 7 9
32    0 8 1
33    0 8 3
34    0 8 5
35    0 8 7
36    0 8 9
37    0 9 1
38    0 9 3
39    0 9 5
40    0 9 7
41    1 0 3
42    1 0 5
43    1 0 7
44    1 0 9
45    1 2 3
46    1 2 5
47    1 2 7
48    1 2 9
49    1 3 5
50    1 3 7
51    1 3 9
52    1 4 3
53    1 4 5
54    1 4 7
55    1 4 9
56    1 5 3
57    1 5 7
58    1 5 9
59    1 6 3
60    1 6 5
61    1 6 7
62    1 6 9
63    1 7 3
64    1 7 5
65    1 7 9
66    1 8 3
67    1 8 5
68    1 8 7
69    1 8 9
70    1 9 3
71    1 9 5
72    1 9 7
73    2 0 1
74    2 0 3
75    2 0 5
76    2 0 7
77    2 0 9
78    2 1 3
79    2 1 5
80    2 1 7
81    2 1 9
82    2 3 1
83    2 3 5
84    2 3 7
85    2 3 9
86    2 4 1
87    2 4 3
88    2 4 5
89    2 4 7
90    2 4 9
91    2 5 1
92    2 5 3
93    2 5 7
94    2 5 9
95    2 6 1
96    2 6 3
97    2 6 5
98    2 6 7
99    2 6 9
100    2 7 1
101    2 7 3
102    2 7 5
103    2 7 9
104    2 8 1
105    2 8 3
106    2 8 5
107    2 8 7
108    2 8 9
109    2 9 1
110    2 9 3
111    2 9 5
112    2 9 7
113    3 0 1
114    3 0 5
115    3 0 7
116    3 0 9
117    3 1 5
118    3 1 7
119    3 1 9
120    3 2 1
121    3 2 5
122    3 2 7
123    3 2 9
124    3 4 1
125    3 4 5
126    3 4 7
127    3 4 9
128    3 5 1
129    3 5 7
130    3 5 9
131    3 6 1
132    3 6 5
133    3 6 7
134    3 6 9
135    3 7 1
136    3 7 5
137    3 7 9
138    3 8 1
139    3 8 5
140    3 8 7
141    3 8 9
142    3 9 1
143    3 9 5
144    3 9 7
. . .

 

Answer 3

Break down the different ways in which 3 digit odd numbers can be formed:

 

1) All 3 digits are odd (e.g. 135), thus there are 5.4.3 =  60 such numbers

 

2) odd, even, odd (e.g. 123), there are 5.5.4 = 100 such numbers

 

3) even, odd, odd (e.g. 215) there are 4.5.4 =80 such numbers ( no Zeroes at the beginning)

 

4) even, even, odd (e.g. 201) there are 4.4.5 = 80 such numbers

 

Adds up to 320.

 

Now for practice answer the same question with an additional condition

that the last digit has to be the largest of all the 3 digits.

 

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Answer 4

You have a 3 digit odd number, so let's assign a blank for each number: _ _ _ . 

 

The basic question is: How many possibilities are there for each blank? Well, there are 5 odd digits: 1, 3, 5, 7, and 9. So there are 5 possibilities for the first blank:   5 _ _

 

The second blank can't have the same digit as the first one, whatever it was. So there are only 4 possible digits for the second blank:   5 4 _. 

 

The third blank can't have the same digit as the other two blanks, so there are only 3 possible digits for it:   5 4 3. 

 

Multiply the blanks' possibilities, and you'll get the answer. 5*4*3 = 60.