A box contains three pairs of blue gloves and two pairs of

The Blue pairs are not distinct. So BLeft of all 3 are same. So are BRight of all 3.
When you do 5C1, you are picking one of the 5 pairs in 5 ways. But picking one Blue pair is the same as picking the other Blue pair.

Responding to a pm:

Yes, 10C1*9C1*8C1 implies that we have 10 distinct objects and we have to pick 3 of them and arrange them in 3 distinct slots: first place, second place and third place.
But we don't have any distinct slots. This is a combinations problem. We need to just select, not arrange in any slots.

Your approach is correct if you know why you did what you did. My only question is this: Why did you multiply by 3!/2! instead of 3! (different arrangements of BBG) since the two Bs are different?

If you understand that when you say 6/10 * 3/9, you are already counting in all arrangements of Bleft and Bright and now all you need to do is arrange G with respect to the 2 Bs (i.e. multiply by 3 for the 3 spots where we can put G), then absolutely, go ahead. There is nothing wrong.
(When you say 6/10, you are counting the possibilities of picking a Bleft or a Bright first and whatever is leftover next, so you have already arranged the different Bs.)[/quote]

Hi VeritasKarishma, I followed the above method. For getting BBG our prob is 1/10 for GGB it will be 4/10*2/9*6/8 = 2/30

adding both we get 1/10+2/30 = 5/30

why is this wrong??

You have accounted for BBG, but how about BGB and GBB? These will give two more 1/10s.
Also, how about BBB (one pair of Bs and another extra B)
Similarly, GGB and GGG.
Check out the complete solution here: https://gmatclub.com/forum/a-box-contai ... l#p1045908

Hmmm, I got very close here. Wonder what I missed?

B = Blue, G = Green, L = Left, R = Right


Total cases = 10C3 = 120

BL + BR can be combined with GR + GL in (2 + 2)*6 ways and with BL + BR in (2 + 2)*3 ways = 36 ways

GL + GR can be combined with BL + BR in (3 + 3)*6 ways and with GR + GL in (1 + 1)*3 ways = 42 ways

In total 78/120 or 39/60

Somewhere I miss 4 cases.

Question: A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?
(A) 3/10 (B) 23/60 (C) 7/12 (D) 41/60 (E) 5/6

BL BR and (BL / BR / GL / GR): 54
(BL BR) BL = 3c1 x 3c1 x 2c1/2! = 9
(BL BR) BR = 3c1 x 3c1 x 2c1/2! = 9
(BL BR) GL = 3c1 x 3c1 x 2c1 = 18
(BL BR) GR = 3c1 x 3c1 x 2c1 = 18

GL GR and (BL / BR / GL / GR) : 28
(GL GR) BL = 2c1 x 2c1 x 3c1 = 12
(GL GR) BR = 2c1 x 2c1 x 3c1 = 12
(GL GR) GL = 2c1 x 2c1 x 1c1/2! = 2
(GL GR) GR = 2c1 x 2c1 x 1c1/2! = 2

Total = 54 + 28 = 82
Probability = 82/10c3 = 82/120 = 41/60

Hope this helps!

According to your logic here, Bright gloves are all identical to each other and the same goes for Bleft, Gright, and Gleft. (This logic is being used to calculate the arrangements in the above solution.)
But if this is the assumption,
then the number of ways of selecting Bleft, Bright, G is 1 not 3*3*4 and the same goes for the remaining cases as well. 
Because the number of ways of selecting r items from n identical items is 1 not nCr. And The number of ways of selecting 1 item of type1, 1 item of type2, and 1 item of type3 from m identical items of type1, n identical items of type2, and l identical items of type3 is equal to 1 as well.
Could you please let me know of any errors in the above reasoning?

We have

• 3 blue left
• 3 blue right
• 2 green left
• 2 green right­

1) 1 Blue pair + 1 Green: 3 * 3 * 4 = 36

2) All 3 are Blue

3) 1 Green pair + 1 Blue: 2 * 2 * 6 = 24

4) All 3 are Green
=> in any case we will get 1 pair
=> 4C3 = 4


==> Total number of possible cases is 36 + 18 + 24 + 4 = 82

Total number of cases is 10C3 = 120

Probability = \(\frac{82}{120} = \frac{41}{60}\)­

­Hi,

Could someone please highlight the flaw in my approach :
6/10 (Selecting one glove out of 6 Blue gloves)* 3/9 (Selecting the other pair of Blue glove)* 8/8 (Selecting any other glove because it does not matter now, since we have one complete pair)

+

4/10*2/9*8/8 (Same approach as above for Green gloves)

= 13/45? What am I missing here? 

something is bugging me here, when we say 10C3 we are considering all the arrangements that differ at least by one element, not the order. so by doing 10C3 we are assuming the 10 elements are different from each other and thus like
BR1 BR2 BR3
BL1 BL2 BL3
GR1 GR2
GL1 GL2

so when we take an arrangement like {BR1 BR2 BL3} and {BR1 BR3 BL3}, both are of the type {BR BR BL} but we are effectively counting them BOTH among the 120 combinations. so i don't understand why i cannot solve the question by doing the following:

3*3*8 = 72 (we can get BR in 3 ways, BL in 3 ways and fill the gap in 8 ways)
and
2*2*8 = 32 (we can get GR in 2 ways, GL in 2 ways and fill the gap in 8 ways)

72+32 should be all the arrangements in 120 that satisfy the condition.

i do not see where the double counting is.
KarishmaB
Bunuel

can anyone explain?­

­
I believe your doubt is addressed here:

https://gmatclub.com/forum/a-box-contai ... l#p1279980

2 Blues, 1 Green:
3C1*3C1*4C1 = 36

2 Greens, 1 Blue
2C1*2C1*6C1 = 24


Hi Karishma,

Can you please help understand why haven't you multiplied the above by 3!/2

Bunuel KarishmaB

What is wrong with this approach ?

i tried doing this as:
(BBB) 3C1*3C1*4C1 + (BBG)3C1*3C1*4C1+(GGG)2C1*2C1*2C1+ (GGB) 2C1*2C1*6C1 divided by 10C3

Please explain where is the fault?

It is luring to go for 3C1 x 3C1 x 8C1 + 2C1 x 2C1 x 8C1 = 104
But many repetitions happen in these 104 outcomes.

For example, in the 1st case, let's assume all 3 are blue.
So, L1, L2, L3, R1, R2, R3 are the total options to choose from.
3C1 x 3C1 = (L1 R1), (L1 R2), (L1 R3), (L2 R1), (L2 R2), (L2 R3), (L3 R1), (L3 R2), (L3 R3) = 9 ways.

Now, say, the 3rd one is a Blue left glove. In the above formula, out of total 8 in 8C1, 2 Blue left, 2 Blue right, and 4 Greens remain. So, when we multiply 3C1 x 3C1 x 2C1 (for the 2 Blue left) we get 18 ways (we are not multiplying by 8C1 because we only want outcomes that have 2 Blue left as the 3rd glove):
(L1 R1 L2), (L1 R1 L3),
(L1 R2 L2), (L1 R2 L3),
(L1 R3 L2), (L1 R3 L3),
(L2 R1 L1), (L2 R1 L3),
(L2 R2 L1), (L2 R2 L3),
(L2 R3 L1), (L2 R3 L3),
(L3 R1 L1), (L3 R1 L2),
(L3 R2 L1), (L3 R2 L2),
(L3 R3 L1), (L3 R3 L2).

Upon looking closely, we can see that 1 and 7 are similar, 2 and 13 are similar, and so on. All are counted twice as ORDER DOESN’T MATTER. Similar double counting is possible when we multiply by 2C1 for 2 Blue right as the 3rd glove.

So, to correct this, we should break up our cases:
Case 1: 2 Blue left, 1 Blue right = 3C2 x 3C1 = 9 [No double counting]
Case 2: 1 Blue left, 2 Blue right = 3C1 x 3C2 = 9 [No double counting]
Case 3: 1 Blue left, 1 Blue right, 1 Green = 3C1 x 3C1 x 4C1 = 36
Case 4: 2 Green left, 1 Green right = 2C2 x 2C1 = 2 [No double counting]
Case 5: 1 Green left, 2 Green right = 2C1 x 2C2 = 2 [No double counting]
Case 6: 1 Green left, 1 Green right, 1 Blue = 2C1 x 2C1 x 6C1 = 24

Total favourable ways = 9 + 9 + 36 + 2 + 2 + 24 = 82
Total possible ways to select 3 gloves from 10 (5 pair) = 10C3 = 120
Probability = 82/120 = 41/60

In combinations, while counting, we should break up our cases so that similar type of items is selected together instead of in multiple separate instances. Otherwise, we need to be cognizant of the double counting and divide the outcome accordingly.

The problem with this approach is that a given glove can be both in the first draw AND in the draw from the group of remaining gloves, leading to double counting.

For example, embedded in the possibilities following this approach is:

3 ways to pick a blue left * 2 ways to pick a blue left from the draw from the remaining 8 gloves = 6 ways.

But the proper approach would be to be to treat this as picking 2 blue gloves from 3 or 3C2 = 3 ways.

So the approach of picking from the remaining gloves doesn't work because a given glove could be both in the first pick as well as in the group of remaining gloves.

why would we not do the same for ggg