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Let $A=\{10,11,12,13, \dots ,99\}$. How many pairs of numbers $x$ and $y$ are possible so that $x+y\geq 100$ and $x$ and $y$ belong to $A$?

  1. $2405$
  2. $2455$
  3. $1200$
  4. $1230$ 
in Combinatory by Boss (16.8k points)
edited by | 38 views
0
I'm getting B as the answer.
0
how?

I got 3605

$10+11+12+....+79+98+98+....10 times 98=3605$
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@srestha I answered. Will you please check if it looks correct? 

1 Answer

0 votes

Answer - Option (B)

Dividing the set into two sets for convenience without loss of generality. 

$S_1 = \{ 10, 11, 12, 13, ..., 49 \}$ and $S_2 = \{50,51,52,53, ..., 99 \}$

Here, No. of elements in $S_1 = 40$ and No. of elements in $S_2 = 50$ 

Now, to make $x + y \geq 100$

we can choose either any 2 numbers from $S_2$ [Part A] or one number from $S_1$ and one number from $S_2$ [Part B].

Since ordered pairs are not asked, pairs $(a,b)$ and $(b,a)$ will be same, hence

 

[Part A]

Total no. of choices = ways of selecting $2$ distinct numbers from set $S_2$= $^{50} C_2 = \frac{50 \times 49}{2}$

but the choices like $(50,50)$, $51,51$ are missing above. So, no. of choices such that $x$ and $y$ are equal are $= 50$ 

[Part B]

Now this part is tricky. We have to choose $1$ number from $S_1$ and $1$ number from $S_2$. 

For numbers $\{90, 91, 92, ..., 99 \}$ we can choose any number from $S_1$ and have $x + y \geq 100$

So, for these $10$ numbers, we have 40 choices.

So, total no. of choices  $ = 10 \times 40 = 400$

Now for $89$, we can choose everything from $S_1$ except number $10$ since that would fail the requirement of $x+y \geq 100$.

So,

for $89$ we have $40-1 = 39$ choices 

Similarly, for $88$ we have $40 -2 = 38$ choices since $\{10, 11 \}$ from $S_1$ will fail our criteria of sum greater than 100. 

Similarly, for $87$, we have $40-3 = 37$ choices

...

Similarly, for $51$, we have $1$ choice (which is $49$)

Similarly, for $50$, we have $0$ choices.

Hence total no. of choices $ = 0 + 1 + 2+ 3+ 4+ ... + 39 = \frac{(39)(40)}{2} = 780$  

 

Hence, Answer $= 1225 +50+780 + 400  = 2455$ 

by Active (1.6k points)
edited by
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@toxicdesire

I suppose, there are many possibility which have u missed here. 

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Can you name any? @srestha

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See my calculation, if u can get it
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@srestha is this pair valid ? $(50,50)$ i.e. can I use same element twice ? I have excluded those.  If I can include these, then the answer will be $2455$. 

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No, see $51$ doesnot have $1$ choice, as u said

for $51$, we have $1$ choice

it has 49 choices $50$ to $99$ and then $-1$ to avoid self loop 

0
So, 2455 is the correct answer.
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