Number of numbers - P&C
$begingroup$
Consider all the $5$ digit numbers where each of the digits is chosen from the set ${1,2,3,4}$. Then the number of numbers which contain all the $4$ digits is?
I don't know where to start. Would love some help/hints.
combinatorics permutations combinations
edited Dec 22 '18 at 12:22
Ankit Kumar
1,514221
asked Dec 22 '18 at 12:05
user3508140user3508140
14438
$endgroup$
add a comment |
$begingroup$
Consider all the $5$ digit numbers where each of the digits is chosen from the set ${1,2,3,4}$. Then the number of numbers which contain all the $4$ digits is?
I don't know where to start. Would love some help/hints.
combinatorics permutations combinations
edited Dec 22 '18 at 12:22
Ankit Kumar
1,514221
asked Dec 22 '18 at 12:05
user3508140user3508140
14438
$endgroup$
add a comment |
$begingroup$
Consider all the $5$ digit numbers where each of the digits is chosen from the set ${1,2,3,4}$. Then the number of numbers which contain all the $4$ digits is?
I don't know where to start. Would love some help/hints.
combinatorics permutations combinations
edited Dec 22 '18 at 12:22
Ankit Kumar
1,514221
asked Dec 22 '18 at 12:05
user3508140user3508140
14438
$endgroup$
Consider all the $5$ digit numbers where each of the digits is chosen from the set ${1,2,3,4}$. Then the number of numbers which contain all the $4$ digits is?
I don't know where to start. Would love some help/hints.
combinatorics permutations combinations
combinatorics permutations combinations
edited Dec 22 '18 at 12:22
Ankit Kumar
1,514221
asked Dec 22 '18 at 12:05
user3508140user3508140
14438
edited Dec 22 '18 at 12:22
Ankit Kumar
1,514221
asked Dec 22 '18 at 12:05
user3508140user3508140
14438
edited Dec 22 '18 at 12:22
Ankit Kumar
1,514221
edited Dec 22 '18 at 12:22
Ankit Kumar
1,514221
edited Dec 22 '18 at 12:22
Ankit Kumar
1,514221
1,514221
asked Dec 22 '18 at 12:05
user3508140user3508140
14438
asked Dec 22 '18 at 12:05
user3508140user3508140
14438
asked Dec 22 '18 at 12:05
user3508140user3508140
14438
14438
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
HINT:
Since there are just 4 numbers, and 5 digits and you want all of them to be present, you'll have to take care only of a few things. Note that you'll have to pick only one extra digit. In how many ways can you do that? After choosing that digit, in how many ways can you arrange them?
SOLUTION:
Ways to choose extra digit$=4$
Ways to arrange them$=5!/2$
Answer = Total combinations$=2*5!$
answered Dec 22 '18 at 12:09
Ankit KumarAnkit Kumar
1,514221
$endgroup$
$begingroup$
4c1*5!? This omes out to be 480. But the answer is 240.
$endgroup$
– user3508140
Dec 22 '18 at 12:18
$begingroup$
Nope. It is 240. $5!=120, 2*120=240$.
$endgroup$
– Ankit Kumar
Dec 22 '18 at 12:18
1
$begingroup$
Missed the 2. Yes. Got it. Thanks.
$endgroup$
– user3508140
Dec 22 '18 at 12:20
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT:
Since there are just 4 numbers, and 5 digits and you want all of them to be present, you'll have to take care only of a few things. Note that you'll have to pick only one extra digit. In how many ways can you do that? After choosing that digit, in how many ways can you arrange them?
SOLUTION:
Ways to choose extra digit$=4$
Ways to arrange them$=5!/2$
Answer = Total combinations$=2*5!$
answered Dec 22 '18 at 12:09
Ankit KumarAnkit Kumar
1,514221
$endgroup$
$begingroup$
4c1*5!? This omes out to be 480. But the answer is 240.
$endgroup$
– user3508140
Dec 22 '18 at 12:18
$begingroup$
Nope. It is 240. $5!=120, 2*120=240$.
$endgroup$
– Ankit Kumar
Dec 22 '18 at 12:18
1
$begingroup$
Missed the 2. Yes. Got it. Thanks.
$endgroup$
– user3508140
Dec 22 '18 at 12:20
add a comment |
$begingroup$
HINT:
Since there are just 4 numbers, and 5 digits and you want all of them to be present, you'll have to take care only of a few things. Note that you'll have to pick only one extra digit. In how many ways can you do that? After choosing that digit, in how many ways can you arrange them?
SOLUTION:
Ways to choose extra digit$=4$
Ways to arrange them$=5!/2$
Answer = Total combinations$=2*5!$
answered Dec 22 '18 at 12:09
Ankit KumarAnkit Kumar
1,514221
$endgroup$
$begingroup$
4c1*5!? This omes out to be 480. But the answer is 240.
$endgroup$
– user3508140
Dec 22 '18 at 12:18
$begingroup$
Nope. It is 240. $5!=120, 2*120=240$.
$endgroup$
– Ankit Kumar
Dec 22 '18 at 12:18
1
$begingroup$
Missed the 2. Yes. Got it. Thanks.
$endgroup$
– user3508140
Dec 22 '18 at 12:20
add a comment |
$begingroup$
HINT:
Since there are just 4 numbers, and 5 digits and you want all of them to be present, you'll have to take care only of a few things. Note that you'll have to pick only one extra digit. In how many ways can you do that? After choosing that digit, in how many ways can you arrange them?
SOLUTION:
Ways to choose extra digit$=4$
Ways to arrange them$=5!/2$
Answer = Total combinations$=2*5!$
answered Dec 22 '18 at 12:09
Ankit KumarAnkit Kumar
1,514221
$endgroup$
HINT:
Since there are just 4 numbers, and 5 digits and you want all of them to be present, you'll have to take care only of a few things. Note that you'll have to pick only one extra digit. In how many ways can you do that? After choosing that digit, in how many ways can you arrange them?
SOLUTION:
Ways to choose extra digit$=4$
Ways to arrange them$=5!/2$
Answer = Total combinations$=2*5!$
answered Dec 22 '18 at 12:09
Ankit KumarAnkit Kumar
1,514221
edited Dec 22 '18 at 12:16
answered Dec 22 '18 at 12:09
Ankit KumarAnkit Kumar
1,514221
answered Dec 22 '18 at 12:09
Ankit KumarAnkit Kumar
1,514221
answered Dec 22 '18 at 12:09
Ankit KumarAnkit Kumar
1,514221
1,514221
$begingroup$
4c1*5!? This omes out to be 480. But the answer is 240.
$endgroup$
– user3508140
Dec 22 '18 at 12:18
$begingroup$
Nope. It is 240. $5!=120, 2*120=240$.
$endgroup$
– Ankit Kumar
Dec 22 '18 at 12:18
1
$begingroup$
Missed the 2. Yes. Got it. Thanks.
$endgroup$
– user3508140
Dec 22 '18 at 12:20
add a comment |
$begingroup$
4c1*5!? This omes out to be 480. But the answer is 240.
$endgroup$
– user3508140
Dec 22 '18 at 12:18
$begingroup$
Nope. It is 240. $5!=120, 2*120=240$.
$endgroup$
– Ankit Kumar
Dec 22 '18 at 12:18
1
$begingroup$
Missed the 2. Yes. Got it. Thanks.
$endgroup$
– user3508140
Dec 22 '18 at 12:20
$begingroup$
4c1*5!? This omes out to be 480. But the answer is 240.
$endgroup$
– user3508140
Dec 22 '18 at 12:18
$begingroup$
4c1*5!? This omes out to be 480. But the answer is 240.
$endgroup$
– user3508140
Dec 22 '18 at 12:18
$begingroup$
Nope. It is 240. $5!=120, 2*120=240$.
$endgroup$
– Ankit Kumar
Dec 22 '18 at 12:18
$begingroup$
Nope. It is 240. $5!=120, 2*120=240$.
$endgroup$
– Ankit Kumar
Dec 22 '18 at 12:18
1
1
$begingroup$
Missed the 2. Yes. Got it. Thanks.
$endgroup$
– user3508140
Dec 22 '18 at 12:20
$begingroup$
Missed the 2. Yes. Got it. Thanks.
$endgroup$
– user3508140
Dec 22 '18 at 12:20
add a comment |
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