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# The total number of 4 digit numbers in which the digits are in descending order is:(a) 210(b) 240(c) 320(d) None of the above.  Hint: We have 10 digits in total. They are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. We have to select 4 digits out of 10 digits. The order of selecting does not matter here. That means this is a problem of combination where we have to select 4 digits out of 10 digits and we need to arrange them in descending order.

We know that the combination is a way of selecting items from a collection, such that the order of selection does not matter.
In this problem we need to count the total number of 4 digit numbers in which the digits are in descending order.
That means at first we need to select 4 digits out of 10 digits. We can select them in any order. After selecting we need to arrange them in descending order. Therefore the order of selection really does not matter here. We can select any 4 digits out of 10 in any order.
Therefore, we can apply a combination formula here.

We can select r things from a set of n things where order doesn't matter in ${}^{n}{{C}_{r}}$ ways.
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$
Therefore, we can select 4 digits out of 10 digits in ${}^{10}{{C}_{4}}$ ways.
${}^{10}{{C}_{4}}=\dfrac{10!}{\left( 10-4 \right)!\times 4!}=\dfrac{10\times 9\times 8\times 7\times 6!}{6!\times 1\times 2\times 3\times 4}=210$
After selecting 4 digits, we can arrange those 4 digits only in one way as we have to arrange them in descending order. There is only one way to arrange any number of digits in descending order.
Hence, for each 210 ways we can arrange those 4 digits only in one way.
Therefore, the total number of 4 digit numbers in which the digits are in descending order is 210.
Hence, option (a) is correct.

Note: Here we need to understand that the order of selection does not matter for this problem. That means we have to apply the combination formula, not permutation. In permutation the order of selecting things matters.
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