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[1] The numbers of ways of selecting $3$ pairs from $8$ distinct objects.

[2] The no. of $6$ Digit no. in which exactly $2$ digits are even, must be

[3] The no. of $6$ Digit no. in which exactly $2$ digits are even, must be

$\bf{My\; Try}$ for $(1)$ one::Let we have $8$ distinct objects..

$\left(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7},a_{8}\right)$ and we have to select $3$ pairs

which can be done by $\displaystyle \underbrace{\binom{8}{2}}_{\bf{for\; first\; pair} }\times \underbrace{\binom{6}{2}}_{\bf{for\; second\; pair} }\times \underbrace{\binom{4}{2}}_{\bf{for\; third\; pair} }$

But i did not understand why we divide the above quantity by $3$

Help Required

also for $(2)$ and $(3)$ one

Thanks

juantheron
  • 53,015

1 Answers1

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Suppose that the eight objects are the numbers $1,2,3,4,5,6,7$, and $8$. You might select $\{1,2\}$ as your first pair, $\{4,6\}$ as your second pair, and $\{3,7\}$ as your third pair. You might instead select $\{4,6\}$ as your first pair, $\{3,7\}$ as your second pair, and $\{1,2\}$ as your third pair. Your calculation counts these as two separate outcomes, but of course in both cases you get the same three pairs. In fact you can select these pairs in any of $3!=6$ (not $3$) different orders, so you’re counting this set of $3$ pairs $6$ times. Thus, to get the actual number of different sets of $3$ pairs, you must divide your answer by $3!=6$ to get

$$\frac1{3!}\binom82\binom62\binom42\;.$$

The answer to the second problem depends on whether you’re allowed to have leading zeroes, i.e., on whether $012345$ counts as a $6$-digit number. If it does not, you should split the problem into two cases: the number begins with an odd digit, and the number begins with an even digit.

  • If the first digit is odd, it can be any of the $5$ odd digits ($1,3,5,7,9$). So can each of the other $3$ odd digits. Each of the $2$ even digits can be any of $0,2,4,6$, or $8$, so there are also $5$ choices for each even digit. Thus, there are $5$ choices for each of the $6$ digits. How many ways are there to choose which $2$ of the digits are even?

  • If the first digit is even, it cannot be $0$, so there are only $4$ choices for it. Are there still $5$ choices for each of the other $5$ digits? How many ways are there to choose where the other even digit goes?

The third problem is identical to the second, so it may not be what you meant to ask.

Brian M. Scott
  • 616,228