Permutations with repetitions

How many ways can the digits 1, 2, 2, 3, 3, 3, 4 be permuted into four-digit, three-digit, and two-digit numbers without repetition?
Example:
4 digits: 1223, 2213, 3122, 2313, 4321, etc.
3 digits: 122, 212, 213, 432, etc.
2 digits: 12, 21, 31, 23

Note: I have tried the permutation formula and it gave incorrect results.

Final Answer:

n₄ = 115
n₃ = 43
n₂ = 14

Step-by-step explanation:

a_{1} = 1223; a_{2} = 1224; a_{3} = 1232; a_{4} = 1233; a_{5} = 1234; a_{6} = 1242; a_{7} = 1243; a_{8} = 1322; a_{9} = 1323; a_{10} = 1324; a_{11} = 1332; a_{12} = 1333; a_{13} = 1334; a_{14} = 1342; a_{15} = 1343; a_{16} = 1422; a_{17} = 1423; a_{18} = 1432; a_{19} = 1433; a_{20} = 2123; a_{21} = 2124; a_{22} = 2132; a_{23} = 2133; a_{24} = 2134; a_{25} = 2142; a_{26} = 2143; a_{27} = 2213; a_{28} = 2214; a_{29} = 2231; a_{30} = 2233; a_{31} = 2234; a_{32} = 2241; a_{33} = 2243; a_{34} = 2312; a_{35} = 2313; a_{36} = 2314; a_{37} = 2321; a_{38} = 2323; a_{39} = 2324; a_{40} = 2331; a_{41} = 2332; a_{42} = 2333; a_{43} = 2334; a_{44} = 2341; a_{45} = 2342; a_{46} = 2343; a_{47} = 2412; a_{48} = 2413; a_{49} = 2421; a_{50} = 2423; a_{51} = 2431; a_{52} = 2432; a_{53} = 2433; a_{54} = 3122; a_{55} = 3123; a_{56} = 3124; a_{57} = 3132; a_{58} = 3133; a_{59} = 3134; a_{60} = 3142; a_{61} = 3143; a_{62} = 3212; a_{63} = 3213; a_{64} = 3214; a_{65} = 3221; a_{66} = 3223; a_{67} = 3224; a_{68} = 3231; a_{69} = 3232; a_{70} = 3233; a_{71} = 3234; a_{72} = 3241; a_{73} = 3242; a_{74} = 3243; a_{75} = 3312; a_{76} = 3313; a_{77} = 3314; a_{78} = 3321; a_{79} = 3322; a_{80} = 3323; a_{81} = 3324; a_{82} = 3331; a_{83} = 3332; a_{84} = 3333; a_{85} = 3334; a_{86} = 3341; a_{87} = 3342; a_{88} = 3343; a_{89} = 3412; a_{90} = 3413; a_{91} = 3421; a_{92} = 3422; a_{93} = 3423; a_{94} = 3431; a_{95} = 3432; a_{96} = 3433; a_{97} = 4122; a_{98} = 4123; a_{99} = 4132; a_{100} = 4133; a_{101} = 4212; a_{102} = 4213; a_{103} = 4221; a_{104} = 4223; a_{105} = 4231; a_{106} = 4232; a_{107} = 4233; a_{108} = 4312; a_{109} = 4313; a_{110} = 4321; a_{111} = 4322; a_{112} = 4323; a_{113} = 4331; a_{114} = 4332; a_{115} = 4333; n_{4} = 115

b_{1} = 122; b_{2} = 123; b_{3} = 124; b_{4} = 132; b_{5} = 133; b_{6} = 134; b_{7} = 142; b_{8} = 143; b_{9} = 212; b_{10} = 213; b_{11} = 214; b_{12} = 221; b_{13} = 223; b_{14} = 224; b_{15} = 231; b_{16} = 232; b_{17} = 233; b_{18} = 234; b_{19} = 241; b_{20} = 242; b_{21} = 243; b_{22} = 312; b_{23} = 313; b_{24} = 314; b_{25} = 321; b_{26} = 322; b_{27} = 323; b_{28} = 324; b_{29} = 331; b_{30} = 332; b_{31} = 333; b_{32} = 334; b_{33} = 341; b_{34} = 342; b_{35} = 343; b_{36} = 412; b_{37} = 413; b_{38} = 421; b_{39} = 422; b_{40} = 423; b_{41} = 431; b_{42} = 432; b_{43} = 433; n_{3} = 43

c_{1} = 12; c_{2} = 13; c_{3} = 14; c_{4} = 21; c_{5} = 22; c_{6} = 23; c_{7} = 24; c_{8} = 31; c_{9} = 32; c_{10} = 33; c_{11} = 34; c_{12} = 41; c_{13} = 42; c_{14} = 43; n_{2} = 14

Did you find an error or inaccuracy? Feel free to send us. Thank you!

Showing 2 comments:

Danka

CAN YOU BE MORE SPECIFIC WITH THE ANSWERS?

4 years ago Like

Math student

Thats wrong:

For anyone wondering why you got the wrong answer, here is the real explanation:

To find the number of permutations without repetition for a given set of numbers, we can use the formula for permutations:

P(n, r) = n! / (n - r)!

Where P(n, r) denotes the number of permutations of n items taken r at a time, and n! represents the factorial of n.

Let's calculate the number of permutations for the input 1.2.2.3.3.3.4.

For four digits (r = 4):
n = 7 (total number of digits)
r = 4 (number of digits to be taken at a time)

P(7, 4) = 7! / (7 - 4)!
= 7! / 3!
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1)
= 7 * 6 * 5 * 4
= 840

So, there are 840 permutations of four digits without repetition.

For three digits (r = 3):
n = 7 (total number of digits)
r = 3 (number of digits to be taken at a time)

P(7, 3) = 7! / (7 - 3)!
= 7! / 4!
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / (4 * 3 * 2 * 1)
= 7 * 6 * 5
= 210

So, there are 210 permutations of three digits without repetition.

For two digits (r = 2):
n = 7 (total number of digits)
r = 2 (number of digits to be taken at a time)

P(7, 2) = 7! / (7 - 2)!
= 7! / 5!
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / (5 * 4 * 3 * 2 * 1)
= 7 * 6
= 42

So, there are 42 permutations of two digits without repetition.

2 years ago 1 Like Like

Tips for related online calculators

See also our combinations with repetition calculator.
See also our permutations calculator.
See also our variations calculator.
Would you like to compute the count of combinations?

You need to know the following knowledge to solve this word math problem:

combinatorics

basic operations and concepts

factorial

numbers

natural numbers

Grade of the word problem

high school

Permutations with repetitions

Final Answer:

Step-by-step explanation:

You need to know the following knowledge to solve this word math problem:

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