Mixing
Probability - four random integers between 0-9, that not more than two are the same
Clash Royale CLAN TAG#URR8PPP
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5
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Four integers are chosen at random between 0 and 9, inclusive. Find the probability that (a) not more than 2 are the same.
What I tried: all unique numbers: 63/125,
two same numbers: 72/1000
And then add them both. But the answer in the book is 963/1000. I'm not getting such a high probability. Where have I made the mistake?
probability
edited Sep 4 at 21:54
smci
354211
asked Sep 4 at 9:25
user585380
506
 |Â
show 3 more comments
up vote
5
down vote
favorite
Four integers are chosen at random between 0 and 9, inclusive. Find the probability that (a) not more than 2 are the same.
What I tried: all unique numbers: 63/125,
two same numbers: 72/1000
And then add them both. But the answer in the book is 963/1000. I'm not getting such a high probability. Where have I made the mistake?
probability
edited Sep 4 at 21:54
smci
354211
asked Sep 4 at 9:25
user585380
506
2
Are you sure about the answer in the book? I got $frac9361,000$.
– José Carlos Santos
Sep 4 at 9:32
2
But 63/125 + 72/100 is more than 1...
– TonyK
Sep 4 at 9:37
@JoséCarlosSantos yes re-checked it. Could you please explain how did you get 936/1000 as I'm not even getting close to such a high probability.
– user585380
Sep 4 at 9:38
2
In how many cases we get $4$ distinct integers? Answer: $10times9times8times7$. In how many ways are the first $2$ integers equal but after that you get $2$ new (distinct) integers? Answer: $10times 9times 8$. But this is also the answer to the question: in how many ways are the first and the third ingeres equal and the other $2$ are distinct from each other and the $2$ others? Actually, there are $6$ questions of this sort. Therefore, the answer should be$$frac10times9times8times7+6times10times9times810,000=frac9361,000.$$
– José Carlos Santos
Sep 4 at 9:44
2
@user585380 Sure: $binom42=6$.
– José Carlos Santos
Sep 4 at 9:48
 |Â
show 3 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Four integers are chosen at random between 0 and 9, inclusive. Find the probability that (a) not more than 2 are the same.
What I tried: all unique numbers: 63/125,
two same numbers: 72/1000
And then add them both. But the answer in the book is 963/1000. I'm not getting such a high probability. Where have I made the mistake?
probability
edited Sep 4 at 21:54
smci
354211
asked Sep 4 at 9:25
user585380
506
Four integers are chosen at random between 0 and 9, inclusive. Find the probability that (a) not more than 2 are the same.
What I tried: all unique numbers: 63/125,
two same numbers: 72/1000
And then add them both. But the answer in the book is 963/1000. I'm not getting such a high probability. Where have I made the mistake?
probability
edited Sep 4 at 21:54
smci
354211
asked Sep 4 at 9:25
user585380
506
edited Sep 4 at 21:54
smci
354211
edited Sep 4 at 21:54
smci
354211
edited Sep 4 at 21:54
smci
354211
354211
asked Sep 4 at 9:25
user585380
506
asked Sep 4 at 9:25
user585380
506
asked Sep 4 at 9:25
user585380
506
506
2
Are you sure about the answer in the book? I got $frac9361,000$.
– José Carlos Santos
Sep 4 at 9:32
2
But 63/125 + 72/100 is more than 1...
– TonyK
Sep 4 at 9:37
@JoséCarlosSantos yes re-checked it. Could you please explain how did you get 936/1000 as I'm not even getting close to such a high probability.
– user585380
Sep 4 at 9:38
2
In how many cases we get $4$ distinct integers? Answer: $10times9times8times7$. In how many ways are the first $2$ integers equal but after that you get $2$ new (distinct) integers? Answer: $10times 9times 8$. But this is also the answer to the question: in how many ways are the first and the third ingeres equal and the other $2$ are distinct from each other and the $2$ others? Actually, there are $6$ questions of this sort. Therefore, the answer should be$$frac10times9times8times7+6times10times9times810,000=frac9361,000.$$
– José Carlos Santos
Sep 4 at 9:44
2
@user585380 Sure: $binom42=6$.
– José Carlos Santos
Sep 4 at 9:48
 |Â
show 3 more comments
2
Are you sure about the answer in the book? I got $frac9361,000$.
– José Carlos Santos
Sep 4 at 9:32
2
But 63/125 + 72/100 is more than 1...
– TonyK
Sep 4 at 9:37
@JoséCarlosSantos yes re-checked it. Could you please explain how did you get 936/1000 as I'm not even getting close to such a high probability.
– user585380
Sep 4 at 9:38
2
In how many cases we get $4$ distinct integers? Answer: $10times9times8times7$. In how many ways are the first $2$ integers equal but after that you get $2$ new (distinct) integers? Answer: $10times 9times 8$. But this is also the answer to the question: in how many ways are the first and the third ingeres equal and the other $2$ are distinct from each other and the $2$ others? Actually, there are $6$ questions of this sort. Therefore, the answer should be$$frac10times9times8times7+6times10times9times810,000=frac9361,000.$$
– José Carlos Santos
Sep 4 at 9:44
2
@user585380 Sure: $binom42=6$.
– José Carlos Santos
Sep 4 at 9:48
2
2
Are you sure about the answer in the book? I got $frac9361,000$.
– José Carlos Santos
Sep 4 at 9:32
Are you sure about the answer in the book? I got $frac9361,000$.
– José Carlos Santos
Sep 4 at 9:32
2
2
But 63/125 + 72/100 is more than 1...
– TonyK
Sep 4 at 9:37
But 63/125 + 72/100 is more than 1...
– TonyK
Sep 4 at 9:37
@JoséCarlosSantos yes re-checked it. Could you please explain how did you get 936/1000 as I'm not even getting close to such a high probability.
– user585380
Sep 4 at 9:38
@JoséCarlosSantos yes re-checked it. Could you please explain how did you get 936/1000 as I'm not even getting close to such a high probability.
– user585380
Sep 4 at 9:38
2
2
In how many cases we get $4$ distinct integers? Answer: $10times9times8times7$. In how many ways are the first $2$ integers equal but after that you get $2$ new (distinct) integers? Answer: $10times 9times 8$. But this is also the answer to the question: in how many ways are the first and the third ingeres equal and the other $2$ are distinct from each other and the $2$ others? Actually, there are $6$ questions of this sort. Therefore, the answer should be$$frac10times9times8times7+6times10times9times810,000=frac9361,000.$$
– José Carlos Santos
Sep 4 at 9:44
In how many cases we get $4$ distinct integers? Answer: $10times9times8times7$. In how many ways are the first $2$ integers equal but after that you get $2$ new (distinct) integers? Answer: $10times 9times 8$. But this is also the answer to the question: in how many ways are the first and the third ingeres equal and the other $2$ are distinct from each other and the $2$ others? Actually, there are $6$ questions of this sort. Therefore, the answer should be$$frac10times9times8times7+6times10times9times810,000=frac9361,000.$$
– José Carlos Santos
Sep 4 at 9:44
2
2
@user585380 Sure: $binom42=6$.
– José Carlos Santos
Sep 4 at 9:48
@user585380 Sure: $binom42=6$.
– José Carlos Santos
Sep 4 at 9:48
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
9
down vote
You're missing a factor of $6$ in the second case, since you can choose two slots for the two equal numbers in $binom42$ ways. Then you get $frac9361000$, as José wrote. Since the book says $frac9631000$, apparently they're also counting the case of two pairs, and "more than $2$ are the same" means that there's at least a triple. That adds another $fracbinom42binom10210^4=frac271000$, so the total is then $frac9631000$.
answered Sep 4 at 9:39
joriki
167k10180333
add a comment |Â
up vote
5
down vote
You can solve it by subtraction: the complementary of the event in your problem consists in having a quartet of equal numbers or a triplet and a different number.
The probability of getting four equal numbers is $frac1010000,$ because there are ten numbers available to be all equal.
The probability of having a triplet is $frac4times10times 910000$ because there are four positions where to place the "single one", 10 numbers that the triplet can assume and nine that the single can have.
Summing these two probabilities, you obtain $frac37010000$ which is exactly $1-frac9631000.$
edited Sep 4 at 9:49
joriki
167k10180333
answered Sep 4 at 9:46
Riccardo Ceccon
850220
1
Thanks for the edit :)
– Riccardo Ceccon
Sep 4 at 9:51
add a comment |Â
up vote
5
down vote
You may first calculate the complementary probability.
- There are $10$ 4-digit groups with 4 identical digits
- There are $4 choose 3$ choices of places to put 1 out of 10 digits into these 3 places and 9 other digits to fill the remaining place: $4 choose 3 cdot 10 cdot 9$
All together, the probability of getting 3 or 4 equal digits is
$$P(mbox3 or 4 equal digits) =frac10 + 4 choose 3 cdot 10 cdot 910^4 = frac3710^3 Rightarrow $$
$$P(mboxat most 2 equal) = 1-frac3710^3 = frac9631000 $$
answered Sep 4 at 9:50
trancelocation
5,2251515
Thanks! But could you point out my mistake in doing the normal way without complementary?
– user585380
Sep 4 at 9:56
@user585380: Your first number is right. The second is wrong. But you did not show your reasoning how you got your second number. You may split the second case (2 equal digits) into two subcases: (1) - exactly one pair; (2) - exactly 2 different (!) pairs.
– trancelocation
Sep 4 at 10:35
We can't have 2 different pairs, right? I got the second number by: 9/10*1/10*8/10 (to get two similar numbers).
– user585380
Sep 4 at 14:28
@user585380: Your number $frac910cdot frac110cdot frac810$ corresponds to $frac110$ for the double digit and $frac910$ and $frac810$ for the two remaining digits. But your calculation does not distinguish, for example, between $1123$, $1213$, $1231$, .... Besides this your calculation excludes cases with two pairs like $1122$. I hope this helps you understand what you overlooked in your calculation.
– trancelocation
Sep 5 at 3:41
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
You're missing a factor of $6$ in the second case, since you can choose two slots for the two equal numbers in $binom42$ ways. Then you get $frac9361000$, as José wrote. Since the book says $frac9631000$, apparently they're also counting the case of two pairs, and "more than $2$ are the same" means that there's at least a triple. That adds another $fracbinom42binom10210^4=frac271000$, so the total is then $frac9631000$.
answered Sep 4 at 9:39
joriki
167k10180333
add a comment |Â
up vote
9
down vote
You're missing a factor of $6$ in the second case, since you can choose two slots for the two equal numbers in $binom42$ ways. Then you get $frac9361000$, as José wrote. Since the book says $frac9631000$, apparently they're also counting the case of two pairs, and "more than $2$ are the same" means that there's at least a triple. That adds another $fracbinom42binom10210^4=frac271000$, so the total is then $frac9631000$.
answered Sep 4 at 9:39
joriki
167k10180333
add a comment |Â
up vote
9
down vote
up vote
9
down vote
You're missing a factor of $6$ in the second case, since you can choose two slots for the two equal numbers in $binom42$ ways. Then you get $frac9361000$, as José wrote. Since the book says $frac9631000$, apparently they're also counting the case of two pairs, and "more than $2$ are the same" means that there's at least a triple. That adds another $fracbinom42binom10210^4=frac271000$, so the total is then $frac9631000$.
answered Sep 4 at 9:39
joriki
167k10180333
You're missing a factor of $6$ in the second case, since you can choose two slots for the two equal numbers in $binom42$ ways. Then you get $frac9361000$, as José wrote. Since the book says $frac9631000$, apparently they're also counting the case of two pairs, and "more than $2$ are the same" means that there's at least a triple. That adds another $fracbinom42binom10210^4=frac271000$, so the total is then $frac9631000$.
answered Sep 4 at 9:39
joriki
167k10180333
edited Sep 4 at 9:49
answered Sep 4 at 9:39
joriki
167k10180333
answered Sep 4 at 9:39
joriki
167k10180333
answered Sep 4 at 9:39
joriki
167k10180333
167k10180333
add a comment |Â
add a comment |Â
up vote
5
down vote
You can solve it by subtraction: the complementary of the event in your problem consists in having a quartet of equal numbers or a triplet and a different number.
The probability of getting four equal numbers is $frac1010000,$ because there are ten numbers available to be all equal.
The probability of having a triplet is $frac4times10times 910000$ because there are four positions where to place the "single one", 10 numbers that the triplet can assume and nine that the single can have.
Summing these two probabilities, you obtain $frac37010000$ which is exactly $1-frac9631000.$
edited Sep 4 at 9:49
joriki
167k10180333
answered Sep 4 at 9:46
Riccardo Ceccon
850220
1
Thanks for the edit :)
– Riccardo Ceccon
Sep 4 at 9:51
add a comment |Â
up vote
5
down vote
You can solve it by subtraction: the complementary of the event in your problem consists in having a quartet of equal numbers or a triplet and a different number.
The probability of getting four equal numbers is $frac1010000,$ because there are ten numbers available to be all equal.
The probability of having a triplet is $frac4times10times 910000$ because there are four positions where to place the "single one", 10 numbers that the triplet can assume and nine that the single can have.
Summing these two probabilities, you obtain $frac37010000$ which is exactly $1-frac9631000.$
edited Sep 4 at 9:49
joriki
167k10180333
answered Sep 4 at 9:46
Riccardo Ceccon
850220
1
Thanks for the edit :)
– Riccardo Ceccon
Sep 4 at 9:51
add a comment |Â
up vote
5
down vote
up vote
5
down vote
You can solve it by subtraction: the complementary of the event in your problem consists in having a quartet of equal numbers or a triplet and a different number.
The probability of getting four equal numbers is $frac1010000,$ because there are ten numbers available to be all equal.
The probability of having a triplet is $frac4times10times 910000$ because there are four positions where to place the "single one", 10 numbers that the triplet can assume and nine that the single can have.
Summing these two probabilities, you obtain $frac37010000$ which is exactly $1-frac9631000.$
edited Sep 4 at 9:49
joriki
167k10180333
answered Sep 4 at 9:46
Riccardo Ceccon
850220
You can solve it by subtraction: the complementary of the event in your problem consists in having a quartet of equal numbers or a triplet and a different number.
The probability of getting four equal numbers is $frac1010000,$ because there are ten numbers available to be all equal.
The probability of having a triplet is $frac4times10times 910000$ because there are four positions where to place the "single one", 10 numbers that the triplet can assume and nine that the single can have.
Summing these two probabilities, you obtain $frac37010000$ which is exactly $1-frac9631000.$
edited Sep 4 at 9:49
joriki
167k10180333
answered Sep 4 at 9:46
Riccardo Ceccon
850220
edited Sep 4 at 9:49
joriki
167k10180333
edited Sep 4 at 9:49
joriki
167k10180333
edited Sep 4 at 9:49
joriki
167k10180333
167k10180333
answered Sep 4 at 9:46
Riccardo Ceccon
850220
answered Sep 4 at 9:46
Riccardo Ceccon
850220
answered Sep 4 at 9:46
Riccardo Ceccon
850220
850220
1
Thanks for the edit :)
– Riccardo Ceccon
Sep 4 at 9:51
add a comment |Â
1
Thanks for the edit :)
– Riccardo Ceccon
Sep 4 at 9:51
1
1
Thanks for the edit :)
– Riccardo Ceccon
Sep 4 at 9:51
Thanks for the edit :)
– Riccardo Ceccon
Sep 4 at 9:51
add a comment |Â
up vote
5
down vote
You may first calculate the complementary probability.
- There are $10$ 4-digit groups with 4 identical digits
- There are $4 choose 3$ choices of places to put 1 out of 10 digits into these 3 places and 9 other digits to fill the remaining place: $4 choose 3 cdot 10 cdot 9$
All together, the probability of getting 3 or 4 equal digits is
$$P(mbox3 or 4 equal digits) =frac10 + 4 choose 3 cdot 10 cdot 910^4 = frac3710^3 Rightarrow $$
$$P(mboxat most 2 equal) = 1-frac3710^3 = frac9631000 $$
answered Sep 4 at 9:50
trancelocation
5,2251515
Thanks! But could you point out my mistake in doing the normal way without complementary?
– user585380
Sep 4 at 9:56
@user585380: Your first number is right. The second is wrong. But you did not show your reasoning how you got your second number. You may split the second case (2 equal digits) into two subcases: (1) - exactly one pair; (2) - exactly 2 different (!) pairs.
– trancelocation
Sep 4 at 10:35
We can't have 2 different pairs, right? I got the second number by: 9/10*1/10*8/10 (to get two similar numbers).
– user585380
Sep 4 at 14:28
@user585380: Your number $frac910cdot frac110cdot frac810$ corresponds to $frac110$ for the double digit and $frac910$ and $frac810$ for the two remaining digits. But your calculation does not distinguish, for example, between $1123$, $1213$, $1231$, .... Besides this your calculation excludes cases with two pairs like $1122$. I hope this helps you understand what you overlooked in your calculation.
– trancelocation
Sep 5 at 3:41
add a comment |Â
up vote
5
down vote
You may first calculate the complementary probability.
- There are $10$ 4-digit groups with 4 identical digits
- There are $4 choose 3$ choices of places to put 1 out of 10 digits into these 3 places and 9 other digits to fill the remaining place: $4 choose 3 cdot 10 cdot 9$
All together, the probability of getting 3 or 4 equal digits is
$$P(mbox3 or 4 equal digits) =frac10 + 4 choose 3 cdot 10 cdot 910^4 = frac3710^3 Rightarrow $$
$$P(mboxat most 2 equal) = 1-frac3710^3 = frac9631000 $$
answered Sep 4 at 9:50
trancelocation
5,2251515
Thanks! But could you point out my mistake in doing the normal way without complementary?
– user585380
Sep 4 at 9:56
@user585380: Your first number is right. The second is wrong. But you did not show your reasoning how you got your second number. You may split the second case (2 equal digits) into two subcases: (1) - exactly one pair; (2) - exactly 2 different (!) pairs.
– trancelocation
Sep 4 at 10:35
We can't have 2 different pairs, right? I got the second number by: 9/10*1/10*8/10 (to get two similar numbers).
– user585380
Sep 4 at 14:28
@user585380: Your number $frac910cdot frac110cdot frac810$ corresponds to $frac110$ for the double digit and $frac910$ and $frac810$ for the two remaining digits. But your calculation does not distinguish, for example, between $1123$, $1213$, $1231$, .... Besides this your calculation excludes cases with two pairs like $1122$. I hope this helps you understand what you overlooked in your calculation.
– trancelocation
Sep 5 at 3:41
add a comment |Â
up vote
5
down vote
up vote
5
down vote
You may first calculate the complementary probability.
- There are $10$ 4-digit groups with 4 identical digits
- There are $4 choose 3$ choices of places to put 1 out of 10 digits into these 3 places and 9 other digits to fill the remaining place: $4 choose 3 cdot 10 cdot 9$
All together, the probability of getting 3 or 4 equal digits is
$$P(mbox3 or 4 equal digits) =frac10 + 4 choose 3 cdot 10 cdot 910^4 = frac3710^3 Rightarrow $$
$$P(mboxat most 2 equal) = 1-frac3710^3 = frac9631000 $$
answered Sep 4 at 9:50
trancelocation
5,2251515
You may first calculate the complementary probability.
- There are $10$ 4-digit groups with 4 identical digits
- There are $4 choose 3$ choices of places to put 1 out of 10 digits into these 3 places and 9 other digits to fill the remaining place: $4 choose 3 cdot 10 cdot 9$
All together, the probability of getting 3 or 4 equal digits is
$$P(mbox3 or 4 equal digits) =frac10 + 4 choose 3 cdot 10 cdot 910^4 = frac3710^3 Rightarrow $$
$$P(mboxat most 2 equal) = 1-frac3710^3 = frac9631000 $$
answered Sep 4 at 9:50
trancelocation
5,2251515
answered Sep 4 at 9:50
trancelocation
5,2251515
answered Sep 4 at 9:50
trancelocation
5,2251515
answered Sep 4 at 9:50
trancelocation
5,2251515
5,2251515
Thanks! But could you point out my mistake in doing the normal way without complementary?
– user585380
Sep 4 at 9:56
@user585380: Your first number is right. The second is wrong. But you did not show your reasoning how you got your second number. You may split the second case (2 equal digits) into two subcases: (1) - exactly one pair; (2) - exactly 2 different (!) pairs.
– trancelocation
Sep 4 at 10:35
We can't have 2 different pairs, right? I got the second number by: 9/10*1/10*8/10 (to get two similar numbers).
– user585380
Sep 4 at 14:28
@user585380: Your number $frac910cdot frac110cdot frac810$ corresponds to $frac110$ for the double digit and $frac910$ and $frac810$ for the two remaining digits. But your calculation does not distinguish, for example, between $1123$, $1213$, $1231$, .... Besides this your calculation excludes cases with two pairs like $1122$. I hope this helps you understand what you overlooked in your calculation.
– trancelocation
Sep 5 at 3:41
add a comment |Â
Thanks! But could you point out my mistake in doing the normal way without complementary?
– user585380
Sep 4 at 9:56
@user585380: Your first number is right. The second is wrong. But you did not show your reasoning how you got your second number. You may split the second case (2 equal digits) into two subcases: (1) - exactly one pair; (2) - exactly 2 different (!) pairs.
– trancelocation
Sep 4 at 10:35
We can't have 2 different pairs, right? I got the second number by: 9/10*1/10*8/10 (to get two similar numbers).
– user585380
Sep 4 at 14:28
@user585380: Your number $frac910cdot frac110cdot frac810$ corresponds to $frac110$ for the double digit and $frac910$ and $frac810$ for the two remaining digits. But your calculation does not distinguish, for example, between $1123$, $1213$, $1231$, .... Besides this your calculation excludes cases with two pairs like $1122$. I hope this helps you understand what you overlooked in your calculation.
– trancelocation
Sep 5 at 3:41
Thanks! But could you point out my mistake in doing the normal way without complementary?
– user585380
Sep 4 at 9:56
Thanks! But could you point out my mistake in doing the normal way without complementary?
– user585380
Sep 4 at 9:56
@user585380: Your first number is right. The second is wrong. But you did not show your reasoning how you got your second number. You may split the second case (2 equal digits) into two subcases: (1) - exactly one pair; (2) - exactly 2 different (!) pairs.
– trancelocation
Sep 4 at 10:35
@user585380: Your first number is right. The second is wrong. But you did not show your reasoning how you got your second number. You may split the second case (2 equal digits) into two subcases: (1) - exactly one pair; (2) - exactly 2 different (!) pairs.
– trancelocation
Sep 4 at 10:35
We can't have 2 different pairs, right? I got the second number by: 9/10*1/10*8/10 (to get two similar numbers).
– user585380
Sep 4 at 14:28
We can't have 2 different pairs, right? I got the second number by: 9/10*1/10*8/10 (to get two similar numbers).
– user585380
Sep 4 at 14:28
@user585380: Your number $frac910cdot frac110cdot frac810$ corresponds to $frac110$ for the double digit and $frac910$ and $frac810$ for the two remaining digits. But your calculation does not distinguish, for example, between $1123$, $1213$, $1231$, .... Besides this your calculation excludes cases with two pairs like $1122$. I hope this helps you understand what you overlooked in your calculation.
– trancelocation
Sep 5 at 3:41
@user585380: Your number $frac910cdot frac110cdot frac810$ corresponds to $frac110$ for the double digit and $frac910$ and $frac810$ for the two remaining digits. But your calculation does not distinguish, for example, between $1123$, $1213$, $1231$, .... Besides this your calculation excludes cases with two pairs like $1122$. I hope this helps you understand what you overlooked in your calculation.
– trancelocation
Sep 5 at 3:41
add a comment |Â
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2
Are you sure about the answer in the book? I got $frac9361,000$.
– José Carlos Santos
Sep 4 at 9:32
2
But 63/125 + 72/100 is more than 1...
– TonyK
Sep 4 at 9:37
@JoséCarlosSantos yes re-checked it. Could you please explain how did you get 936/1000 as I'm not even getting close to such a high probability.
– user585380
Sep 4 at 9:38
2
In how many cases we get $4$ distinct integers? Answer: $10times9times8times7$. In how many ways are the first $2$ integers equal but after that you get $2$ new (distinct) integers? Answer: $10times 9times 8$. But this is also the answer to the question: in how many ways are the first and the third ingeres equal and the other $2$ are distinct from each other and the $2$ others? Actually, there are $6$ questions of this sort. Therefore, the answer should be$$frac10times9times8times7+6times10times9times810,000=frac9361,000.$$
– José Carlos Santos
Sep 4 at 9:44
2
@user585380 Sure: $binom42=6$.
– José Carlos Santos
Sep 4 at 9:48